How can I generate a number between 1 and 9 using a function which generate a number between 1 and 3? The probability to obtain any number between 1 and 9 must be the same, so rand3()+rand3()+rand3() is not a good solution.
Try cartesian-like product:
Rand9() = 3 * (Rand3() - 1) + Rand3()
With 3 * (Rand3() - 1) you make sub-intervals 1-3, 4-6 and 7-9 equally likely. With + Rand3() you will then choose equally on that sub-interval.
Written as a product:
3 * (Rand3() - 1) + Rand3() -> {1, 4, 7} X {+1,+2,+3} -> {1,2,3,4,5,6,7,8,9}
The main idea of the solution is to build 3*3 matrix.
[
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
]
function init() {
const rand3 = generateMathRandom(3);
let randFunc = rand3;
const start = 9;
const end = 10;
let i = start;
console.time('---start');
while (i < end) {
randFunc = generateRandom(i, 3, rand3);
healthCheck(i, randFunc);
i++;
}
console.timeEnd('---start');
i = start;
console.time('---start-math');
while (i < end) {
randFunc = generateMathRandom(i);
healthCheck(i, randFunc);
i++;
}
console.timeEnd('---start-math');
}
function generateMathRandom(range) {
return function () {
return Math.floor(Math.random() * range);
}
}
// range - target random range number : e.g 9
// lowerRange - base random range number : e.g 3
// func - function which generate random with the range of lowerRange : e.g rand3
function generateRandom(range, lowerRange, func) {
return function () {
const results = [];
let times = 1;
let matRange = lowerRange;
let mat = Math.sqrt(range);
while(mat > lowerRange) {
mat = Math.sqrt(mat);
times ++;
matRange *= lowerRange;
}
const noneStart = Math.floor(matRange * matRange / range) * range;
for (let i = 0; i < matRange; i++) {
results.push([]);
for (let j = 0; j < matRange; j++) {
const num = i * matRange + j;
if (num < noneStart)
results[i].push(num % range);
else
results[i].push(-1);
}
}
while (true) {
const row = new Array(times).fill(1).map(n => func()).reduce((a, c) => a * lowerRange + c, 0);
const col = new Array(times).fill(1).map(n => func()).reduce((a, c) => a * lowerRange + c, 0);
if (!results[row]) {
debugger;
}
if (results[row][col] == undefined) {
debugger;
}
if (results[row][col] === -1) continue;
return results[row][col];
}
}
}
function healthCheck(range, func = null, count = 100000) {
const funcToCheck = func || rand[range] || null;
if (!funcToCheck) {
console.log('Function is not provided');
return;
}
const health = new Array(range).fill(0);
const checkCount = count - count % range; // To do correct check.
for (let i = 0; i < checkCount; i++) {
health[funcToCheck()]++;
}
const result = health.map((cnt, index) => ({
_value: index,
rate: (cnt / checkCount * 100).toFixed(5)
}));
// console.log('Random result:', result);
let minRate = 100, maxRate = 0;
for (let i = 0; i < range; i++) {
minRate = Math.min(Number(result[i].rate), minRate);
maxRate = Math.max(Number(result[i].rate), maxRate);
}
console.log('Random health<' + range + '>: ' + healthOutput(range, maxRate - minRate));
}
function healthOutput(range, rangeOffset) {
const healthThreshold = 2; // It should be lower for the best Alghorithms
const threshold = healthThreshold;
if (rangeOffset < threshold * 0.2) {
return 'The Super! -> ' + rangeOffset.toFixed(5);
} else if (rangeOffset < threshold * 0.5) {
return 'The Best! --> ' + rangeOffset.toFixed(5);
} else if (rangeOffset < threshold * 1) {
return 'Good work! -> ' + rangeOffset.toFixed(5);
} else if (rangeOffset < threshold * 2) {
return 'Not bad! ---> ' + rangeOffset.toFixed(5);
}
return 'Worst! -----> ' + rangeOffset.toFixed(5);
}
init();
I have a kernel code that executes properly
runnable code
__global__ static void CalcSTLDistance_Kernel(Integer ComputeParticleNumber)
{
//const Integer TID = CudaGetTargetID();
const Integer ID =CudaGetTargetID();
/*if(ID >= ComputeParticleNumber)
{
return ;
}*/
CDistance NearestDistance;
Integer NearestID = -1;
NearestDistance.Magnitude = 1e8;
NearestDistance.Direction.x = 0;
NearestDistance.Direction.y = 0;
NearestDistance.Direction.z = 0;//make_Scalar3(0,0,0);
//if(c_daOutputParticleID[ID] < -1)
//{
// c_daSTLDistance[ID] = NearestDistance;
// c_daSTLID[ID] = NearestID;
// return;
//}
//Scalar3 TargetPosition = c_daParticlePosition[ID];
Integer TriangleID;
Integer CIDX, CIDY, CIDZ;
Integer CID = GetCellID(&CONSTANT_BOUNDINGBOX,&c_daParticlePosition[ID],CIDX, CIDY, CIDZ);
if(CID >=0 && CID < c_CellNum)
{
//Integer Range = 1;
for(Integer k = -1; k <= 1; ++k)
{
for(Integer j = -1; j <= 1; ++j)
{
for(Integer i = -1; i <= 1; ++i)
{
Integer MCID = GetCellID(&CONSTANT_BOUNDINGBOX,CIDX +i, CIDY + j,CIDZ + k);
if(MCID < 0 || MCID >= c_CellNum)
{
continue;
}
unsigned int TriangleNum = c_daCell[MCID].m_TriangleNum;
for(unsigned int l = 0; l < TriangleNum; ++l)
{
TriangleID = c_daCell[MCID].m_TriangleID[l];
/*if(c_daTrianglesParameters[c_daTriangles[TriangleID].ModelIDNumber].isDrag)
{
continue;
}*/
if( TriangleID >= 0 && TriangleID < c_TriangleNum && TriangleID != NearestID)// No need to calculate again for the same triangle
{
CDistance Distance ;
Distance.Magnitude = CalcDistance(&c_daTriangles[TriangleID], &c_daParticlePosition[ID], &Distance.Direction);
if(Distance.Magnitude < NearestDistance.Magnitude)
{
NearestDistance = Distance;
NearestID = TriangleID;
}
}
}
}
}
}
}
c_daSTLDistance[ID] = NearestDistance;
c_daSTLID[ID] = NearestID;
}
and when I add any basic variables or perform any checking operation, it gives unknown error and while checking wih cuda-memcheck, it suggests memory read error.
here in the changed code, i tried to check the previously calculated part and tried to skip the redundant calculation. for this I tried to perform basic check operation in array but it throws memory error.
error raising code
__global__ static void CalcSTLDistance_Kernel(Integer ComputeParticleNumber)
{
//const Integer TID = CudaGetTargetID();
const Integer ID =CudaGetTargetID();
/*if(ID >= ComputeParticleNumber)
{
return ;
}*/
CDistance NearestDistance;
Integer NearestID = -1;
NearestDistance.Magnitude = 1e8;
NearestDistance.Direction.x = 0;
NearestDistance.Direction.y = 0;
NearestDistance.Direction.z = 0;//make_Scalar3(0,0,0);
//if(c_daOutputParticleID[ID] < -1)
//{
// c_daSTLDistance[ID] = NearestDistance;
// c_daSTLID[ID] = NearestID;
// return;
//}
//Scalar3 TargetPosition = c_daParticlePosition[ID];
Integer TriangleID;
Integer CIDX, CIDY, CIDZ;
Integer CID = GetCellID(&CONSTANT_BOUNDINGBOX,&c_daParticlePosition[ID],CIDX, CIDY, CIDZ);
int len=0;
int td[100];
for(int m=0;m<100;m++)
{
td[m]=-1;
}
if(CID >=0 && CID < c_CellNum)
{
//Integer Range = 1;
for(Integer k = -1; k <= 1; ++k)
{
for(Integer j = -1; j <= 1; ++j)
{
for(Integer i = -1; i <= 1; ++i)
{
Integer MCID = GetCellID(&CONSTANT_BOUNDINGBOX,CIDX +i, CIDY + j,CIDZ + k);
if(MCID < 0 || MCID >= c_CellNum)
{
continue;
}
unsigned int TriangleNum = c_daCell[MCID].m_TriangleNum;
bool flag = false;
//len=len+TriangleNum ;
for(unsigned int l = 0; l < TriangleNum; ++l)
{
TriangleID = c_daCell[MCID].m_TriangleID[l];
//tem[l] = c_daCell[MCID].m_TriangleID[l];
for(int m=0;m<100;m++)
{
if(TriangleID ==td[m])
{
flag= true;
}
if(flag == true)
break;
}
if(flag == true)
continue;
else
{
td[len] = TriangleID;
len= len+1;
if( TriangleID >= 0 && TriangleID < c_TriangleNum && TriangleID != NearestID)// No need to calculate again for the same triangle
{
CDistance Distance ;
Distance.Magnitude = CalcDistance(&c_daTriangles[TriangleID], &c_daParticlePosition[ID], &Distance.Direction);
if(Distance.Magnitude < NearestDistance.Magnitude)
{
NearestDistance = Distance;
NearestID = TriangleID;
}
}
}
}
}
}
}
}
c_daSTLDistance[ID] = NearestDistance;
c_daSTLID[ID] = NearestID;
}
this problem arises whenever I tried to add any piece of code,thus I suspects that this block of kernel is not allowing me to add any further code due to memory over use.
is there any memory violation rule per block or thread??
how to find the total memory usuage per kernel ?? is there any way??
Well, I am dealing with sudoku solving algorithm and generation but stuck at rather simple task. I have made the check, whether a number is really fit in the position row-wise and column-wise. But what it is driving me mad is block check, ie, whether the number is really fit in the 3x3 block.
It must be simple enough but I can't really arrive at the solution. In short, I want to know the 3x3 block to which a position in matrix belongs. Here are some of the assert cases. The block no, row no and col no indexing starts from 0.
assert("x( 0, 8 ) === 2");
assert("x( 8, 8 ) === 8");
assert("x( 3, 3 ) === 4");
assert("x( 3, 7 ) === 5");
assert("x( 7, 1 ) === 6");
x( i , j ) returns the block number where i = row and j = col.
Isn't it just:
block = 3 * (i / 3) + (j / 3)
(assumes integer operations).
I would code a check, something like this (in pseudo C++)
// row = row to check
// col = column to check
// checkNum = number we are thinking of inserting
bool check(int row, int col, int checkNum)
{
int blockRow = 3 * (row/3);
int blockCol = 3 * (col/3);
for(int i = 0 ; i < 9 ; i++)
{
if(grid[row][i] == checkNum) return false; // number exists in the row.
if(grid[i][col] == checkNum) return false; // number exists in the col.
if(grid[blockRow + i/3][blockCol + i%3] == checkNum) return false; // number exists in the block.
}
return true;
}
Here is a sudoku solver in javascript. Taken from DSSudokuSolver, that I created.
The CleanElements function does something similar to what you are asking for.
CleanElements = function(comp_ary, Qsudoku){
for(i=0; i<9; i++){
for(j=0; j<9; j++){
/*if(Qsudoku[i][j] != ""){
comp_ary[i][j]=[];
}*/
for(k=0; k<9; k++){
i_index = comp_ary[i][k].indexOf(Qsudoku[i][j]);
if(i_index != -1){
comp_ary[i][k].splice(i_index, 1);
}
j_index = comp_ary[k][j].indexOf(Qsudoku[i][j]);
if(j_index != -1){
comp_ary[k][j].splice(j_index, 1);
}
}
if(i < 3){
i_min = 0;
i_max = 2;
}
else if(i < 6){
i_min = 3;
i_max = 5;
}
else{
i_min = 6;
i_max = 8;
}
if(j < 3){
j_min = 0;
j_max = 2;
}
else if(j < 6){
j_min = 3;
j_max = 5;
}
else{
j_min = 6;
j_max = 8;
}
for(i_box=i_min; i_box<=i_max; i_box++){
for(j_box=j_min; j_box<=j_max; j_box++){
index = comp_ary[i_box][j_box].indexOf(Qsudoku[i][j]);
if(index != -1){
comp_ary[i_box][j_box].splice(index, 1);
}
}
}
}
}
return comp_ary;
}
FindElements = function(comp_ary, Qsudoku){
for(i=0; i<9; i++){
for(j=0; j<9; j++){
if(comp_ary[i][j].length == 1){
if (Qsudoku[i][j] == ""){
Qsudoku[i][j] = comp_ary[i][j][0];
comp_ary[i][j] = [];
}
}
}
}
return Qsudoku;
}
IsThereNullElement = function(Qsudoku){
for(i=0; i<9; i++){
for(j=0; j<9; j++){
if(Qsudoku[i][j] == ""){
return false;
}
}
}
return true;
}
InitEmptyArray = function(){
empty_ary = Array();
for(i=0; i<9; i++){
empty_ary[i] = Array();
for(j=0; j<9; j++){
empty_ary[i][j] = Array();
for(k=0; k<9; k++){
empty_ary[i][j][k] = (k+1).toString();
}
}
}
return empty_ary;
}
DSSolve = function(Qsudoku){
comp_ary = InitEmptyArray(); //Complementary Array
window.comp_ary_old = comp_ary;
IterationMax = 5000;
while(true){
IterationMax -= 1;
comp_ary = CleanElements(comp_ary, Qsudoku);
console.log(comp_ary);
if(window.comp_ary_old == comp_ary){
//implement this.
}
else{
window.comp_ary_old = comp_ary;
}
Qsudoku = FindElements(comp_ary, Qsudoku);
//console.log(Qsudoku);
if(IsThereNullElement(Qsudoku)){
return Qsudoku;
}
if(IterationMax == 0){
return null;
}
}
}
I have this string s1 = "My name is X Y Z" and I want to reverse the order of the words so that s1 = "Z Y X is name My".
I can do it using an additional array. I thought hard but is it possible to do it inplace (without using additional data structures) and with the time complexity being O(n)?
Reverse the entire string, then reverse the letters of each individual word.
After the first pass the string will be
s1 = "Z Y X si eman yM"
and after the second pass it will be
s1 = "Z Y X is name My"
reverse the string and then, in a second pass, reverse each word...
in c#, completely in-place without additional arrays:
static char[] ReverseAllWords(char[] in_text)
{
int lindex = 0;
int rindex = in_text.Length - 1;
if (rindex > 1)
{
//reverse complete phrase
in_text = ReverseString(in_text, 0, rindex);
//reverse each word in resultant reversed phrase
for (rindex = 0; rindex <= in_text.Length; rindex++)
{
if (rindex == in_text.Length || in_text[rindex] == ' ')
{
in_text = ReverseString(in_text, lindex, rindex - 1);
lindex = rindex + 1;
}
}
}
return in_text;
}
static char[] ReverseString(char[] intext, int lindex, int rindex)
{
char tempc;
while (lindex < rindex)
{
tempc = intext[lindex];
intext[lindex++] = intext[rindex];
intext[rindex--] = tempc;
}
return intext;
}
Not exactly in place, but anyway: Python:
>>> a = "These pretzels are making me thirsty"
>>> " ".join(a.split()[::-1])
'thirsty me making are pretzels These'
In Smalltalk:
'These pretzels are making me thirsty' subStrings reduce: [:a :b| b, ' ', a]
I know noone cares about Smalltalk, but it's so beautiful to me.
You cannot do the reversal without at least some extra data structure. I think the smallest structure would be a single character as a buffer while you swap letters. It can still be considered "in place", but it's not completely "extra data structure free".
Below is code implementing what Bill the Lizard describes:
string words = "this is a test";
// Reverse the entire string
for(int i = 0; i < strlen(words) / 2; ++i) {
char temp = words[i];
words[i] = words[strlen(words) - i];
words[strlen(words) - i] = temp;
}
// Reverse each word
for(int i = 0; i < strlen(words); ++i) {
int wordstart = -1;
int wordend = -1;
if(words[i] != ' ') {
wordstart = i;
for(int j = wordstart; j < strlen(words); ++j) {
if(words[j] == ' ') {
wordend = j - 1;
break;
}
}
if(wordend == -1)
wordend = strlen(words);
for(int j = wordstart ; j <= (wordend + wordstart) / 2 ; ++j) {
char temp = words[j];
words[j] = words[wordend - (j - wordstart)];
words[wordend - (j - wordstart)] = temp;
}
i = wordend;
}
}
What language?
If PHP, you can explode on space, then pass the result to array_reverse.
If its not PHP, you'll have to do something slightly more complex like:
words = aString.split(" ");
for (i = 0; i < words.length; i++) {
words[i] = words[words.length-i];
}
public static String ReverseString(String str)
{
int word_length = 0;
String result = "";
for (int i=0; i<str.Length; i++)
{
if (str[i] == ' ')
{
result = " " + result;
word_length = 0;
} else
{
result = result.Insert(word_length, str[i].ToString());
word_length++;
}
}
return result;
}
This is C# code.
In Python...
ip = "My name is X Y Z"
words = ip.split()
words.reverse()
print ' '.join(words)
Anyway cookamunga provided good inline solution using python!
This is assuming all words are separated by spaces:
#include <stdio.h>
#include <string.h>
int main()
{
char string[] = "What are you looking at";
int i, n = strlen(string);
int tail = n-1;
for(i=n-1;i>=0;i--)
{
if(string[i] == ' ' || i == 0)
{
int cursor = (i==0? i: i+1);
while(cursor <= tail)
printf("%c", string[cursor++]);
printf(" ");
tail = i-1;
}
}
return 0;
}
class Program
{
static void Main(string[] args)
{
string s1 =" My Name varma:;
string[] arr = s1.Split(' ');
Array.Reverse(arr);
string str = string.Join(" ", arr);
Console.WriteLine(str);
Console.ReadLine();
}
}
This is not perfect but it works for me right now. I don't know if it has O(n) running time btw (still studying it ^^) but it uses one additional array to fulfill the task.
It is probably not the best answer to your problem because i use a dest string to save the reversed version instead of replacing each words in the source string. The problem is that i use a local stack variable named buf to copy all the words in and i can not copy but into the source string as this would lead to a crash if the source string is const char * type.
But it was my first attempt to write s.th. like this :) Ok enough blablub. here is code:
#include <iostream>
using namespace std;
void reverse(char *des, char * const s);
int main (int argc, const char * argv[])
{
char* s = (char*)"reservered. rights All Saints. The 2011 (c) Copyright 11/10/11 on Pfundstein Markus by Created";
char *x = (char*)"Dogfish! White-spotted Shark, Bullhead";
printf("Before: |%s|\n", x);
printf("Before: |%s|\n", s);
char *d = (char*)malloc((strlen(s)+1)*sizeof(char));
char *i = (char*)malloc((strlen(x)+1)*sizeof(char));
reverse(d,s);
reverse(i,x);
printf("After: |%s|\n", i);
printf("After: |%s|\n", d);
free (i);
free (d);
return 0;
}
void reverse(char *dest, char *const s) {
// create a temporary pointer
if (strlen(s)==0) return;
unsigned long offset = strlen(s)+1;
char *buf = (char*)malloc((offset)*sizeof(char));
memset(buf, 0, offset);
char *p;
// iterate from end to begin and count how much words we have
for (unsigned long i = offset; i != 0; i--) {
p = s+i;
// if we discover a whitespace we know that we have a whole word
if (*p == ' ' || *p == '\0') {
// we increment the counter
if (*p != '\0') {
// we write the word into the buffer
++p;
int d = (int)(strlen(p)-strlen(buf));
strncat(buf, p, d);
strcat(buf, " ");
}
}
}
// copy the last word
p -= 1;
int d = (int)(strlen(p)-strlen(buf));
strncat(buf, p, d);
strcat(buf, "\0");
// copy stuff to destination string
for (int i = 0; i < offset; ++i) {
*(dest+i)=*(buf+i);
}
free(buf);
}
We can insert the string in a stack and when we extract the words, they will be in reverse order.
void ReverseWords(char Arr[])
{
std::stack<std::string> s;
char *str;
int length = strlen(Arr);
str = new char[length+1];
std::string ReversedArr;
str = strtok(Arr," ");
while(str!= NULL)
{
s.push(str);
str = strtok(NULL," ");
}
while(!s.empty())
{
ReversedArr = s.top();
cout << " " << ReversedArr;
s.pop();
}
}
This quick program works..not checks the corner cases though.
#include <stdio.h>
#include <stdlib.h>
struct node
{
char word[50];
struct node *next;
};
struct stack
{
struct node *top;
};
void print (struct stack *stk);
void func (struct stack **stk, char *str);
main()
{
struct stack *stk = NULL;
char string[500] = "the sun is yellow and the sky is blue";
printf("\n%s\n", string);
func (&stk, string);
print (stk);
}
void func (struct stack **stk, char *str)
{
char *p1 = str;
struct node *new = NULL, *list = NULL;
int i, j;
if (*stk == NULL)
{
*stk = (struct stack*)malloc(sizeof(struct stack));
if (*stk == NULL)
printf("\n####### stack is not allocated #####\n");
(*stk)->top = NULL;
}
i = 0;
while (*(p1+i) != '\0')
{
if (*(p1+i) != ' ')
{
new = (struct node*)malloc(sizeof(struct node));
if (new == NULL)
printf("\n####### new is not allocated #####\n");
j = 0;
while (*(p1+i) != ' ' && *(p1+i) != '\0')
{
new->word[j] = *(p1 + i);
i++;
j++;
}
new->word[j++] = ' ';
new->word[j] = '\0';
new->next = (*stk)->top;
(*stk)->top = new;
}
i++;
}
}
void print (struct stack *stk)
{
struct node *tmp = stk->top;
int i;
while (tmp != NULL)
{
i = 0;
while (tmp->word[i] != '\0')
{
printf ("%c" , tmp->word[i]);
i++;
}
tmp = tmp->next;
}
printf("\n");
}
Most of these answers fail to account for leading and/or trailing spaces in the input string. Consider the case of str=" Hello world"... The simple algo of reversing the whole string and reversing individual words winds up flipping delimiters resulting in f(str) == "world Hello ".
The OP said "I want to reverse the order of the words" and did not mention that leading and trailing spaces should also be flipped! So, although there are a ton of answers already, I'll provide a [hopefully] more correct one in C++:
#include <string>
#include <algorithm>
void strReverseWords_inPlace(std::string &str)
{
const char delim = ' ';
std::string::iterator w_begin, w_end;
if (str.size() == 0)
return;
w_begin = str.begin();
w_end = str.begin();
while (w_begin != str.end()) {
if (w_end == str.end() || *w_end == delim) {
if (w_begin != w_end)
std::reverse(w_begin, w_end);
if (w_end == str.end())
break;
else
w_begin = ++w_end;
} else {
++w_end;
}
}
// instead of reversing str.begin() to str.end(), use two iterators that
// ...represent the *logical* begin and end, ignoring leading/traling delims
std::string::iterator str_begin = str.begin(), str_end = str.end();
while (str_begin != str_end && *str_begin == delim)
++str_begin;
--str_end;
while (str_end != str_begin && *str_end == delim)
--str_end;
++str_end;
std::reverse(str_begin, str_end);
}
My version of using stack:
public class Solution {
public String reverseWords(String s) {
StringBuilder sb = new StringBuilder();
String ns= s.trim();
Stack<Character> reverse = new Stack<Character>();
boolean hadspace=false;
//first pass
for (int i=0; i< ns.length();i++){
char c = ns.charAt(i);
if (c==' '){
if (!hadspace){
reverse.push(c);
hadspace=true;
}
}else{
hadspace=false;
reverse.push(c);
}
}
Stack<Character> t = new Stack<Character>();
while (!reverse.empty()){
char temp =reverse.pop();
if(temp==' '){
//get the stack content out append to StringBuilder
while (!t.empty()){
char c =t.pop();
sb.append(c);
}
sb.append(' ');
}else{
//push to stack
t.push(temp);
}
}
while (!t.empty()){
char c =t.pop();
sb.append(c);
}
return sb.toString();
}
}
Store Each word as a string in array then print from end
public void rev2() {
String str = "my name is ABCD";
String A[] = str.split(" ");
for (int i = A.length - 1; i >= 0; i--) {
if (i != 0) {
System.out.print(A[i] + " ");
} else {
System.out.print(A[i]);
}
}
}
In Python, if you can't use [::-1] or reversed(), here is the simple way:
def reverse(text):
r_text = text.split(" ")
res = []
for word in range(len(r_text) - 1, -1, -1):
res.append(r_text[word])
return " ".join(res)
print (reverse("Hello World"))
>> World Hello
[Finished in 0.1s]
Printing words in reverse order of a given statement using C#:
void ReverseWords(string str)
{
int j = 0;
for (int i = (str.Length - 1); i >= 0; i--)
{
if (str[i] == ' ' || i == 0)
{
j = i == 0 ? i : i + 1;
while (j < str.Length && str[j] != ' ')
Console.Write(str[j++]);
Console.Write(' ');
}
}
}
Here is the Java Implementation:
public static String reverseAllWords(String given_string)
{
if(given_string == null || given_string.isBlank())
return given_string;
char[] str = given_string.toCharArray();
int start = 0;
// Reverse the entire string
reverseString(str, start, given_string.length() - 1);
// Reverse the letters of each individual word
for(int end = 0; end <= given_string.length(); end++)
{
if(end == given_string.length() || str[end] == ' ')
{
reverseString(str, start, end-1);
start = end + 1;
}
}
return new String(str);
}
// In-place reverse string method
public static void reverseString(char[] str, int start, int end)
{
while(start < end)
{
char temp = str[start];
str[start++] = str[end];
str[end--] = temp;
}
}
Actually, the first answer:
words = aString.split(" ");
for (i = 0; i < words.length; i++) {
words[i] = words[words.length-i];
}
does not work because it undoes in the second half of the loop the work it did in the first half. So, i < words.length/2 would work, but a clearer example is this:
words = aString.split(" "); // make up a list
i = 0; j = words.length - 1; // find the first and last elements
while (i < j) {
temp = words[i]; words[i] = words[j]; words[j] = temp; //i.e. swap the elements
i++;
j--;
}
Note: I am not familiar with the PHP syntax, and I have guessed incrementer and decrementer syntax since it seems to be similar to Perl.
How about ...
var words = "My name is X Y Z";
var wr = String.Join( " ", words.Split(' ').Reverse().ToArray() );
I guess that's not in-line tho.
In c, this is how you might do it, O(N) and only using O(1) data structures (i.e. a char).
#include<stdio.h>
#include<stdlib.h>
main(){
char* a = malloc(1000);
fscanf(stdin, "%[^\0\n]", a);
int x = 0, y;
while(a[x]!='\0')
{
if (a[x]==' ' || a[x]=='\n')
{
x++;
}
else
{
y=x;
while(a[y]!='\0' && a[y]!=' ' && a[y]!='\n')
{
y++;
}
int z=y;
while(x<y)
{
y--;
char c=a[x];a[x]=a[y];a[y]=c;
x++;
}
x=z;
}
}
fprintf(stdout,a);
return 0;
}
It can be done more simple using sscanf:
void revertWords(char *s);
void revertString(char *s, int start, int n);
void revertWordsInString(char *s);
void revertString(char *s, int start, int end)
{
while(start<end)
{
char temp = s[start];
s[start] = s[end];
s[end]=temp;
start++;
end --;
}
}
void revertWords(char *s)
{
int start = 0;
char *temp = (char *)malloc(strlen(s) + 1);
int numCharacters = 0;
while(sscanf(&s[start], "%s", temp) !=EOF)
{
numCharacters = strlen(temp);
revertString(s, start, start+numCharacters -1);
start = start+numCharacters + 1;
if(s[start-1] == 0)
return;
}
free (temp);
}
void revertWordsInString(char *s)
{
revertString(s,0, strlen(s)-1);
revertWords(s);
}
int main()
{
char *s= new char [strlen("abc deff gh1 jkl")+1];
strcpy(s,"abc deff gh1 jkl");
revertWordsInString(s);
printf("%s",s);
return 0;
}
import java.util.Scanner;
public class revString {
static char[] str;
public static void main(String[] args) {
//Initialize string
//str = new char[] { 'h', 'e', 'l', 'l', 'o', ' ', 'a', ' ', 'w', 'o',
//'r', 'l', 'd' };
getInput();
// reverse entire string
reverse(0, str.length - 1);
// reverse the words (delimeted by space) back to normal
int i = 0, j = 0;
while (j < str.length) {
if (str[j] == ' ' || j == str.length - 1) {
int m = i;
int n;
//dont include space in the swap.
//(special case is end of line)
if (j == str.length - 1)
n = j;
else
n = j -1;
//reuse reverse
reverse(m, n);
i = j + 1;
}
j++;
}
displayArray();
}
private static void reverse(int i, int j) {
while (i < j) {
char temp;
temp = str[i];
str[i] = str[j];
str[j] = temp;
i++;
j--;
}
}
private static void getInput() {
System.out.print("Enter string to reverse: ");
Scanner scan = new Scanner(System.in);
str = scan.nextLine().trim().toCharArray();
}
private static void displayArray() {
//Print the array
for (int i = 0; i < str.length; i++) {
System.out.print(str[i]);
}
}
}
In Java using an additional String (with StringBuilder):
public static final String reverseWordsWithAdditionalStorage(String string) {
StringBuilder builder = new StringBuilder();
char c = 0;
int index = 0;
int last = string.length();
int length = string.length()-1;
StringBuilder temp = new StringBuilder();
for (int i=length; i>=0; i--) {
c = string.charAt(i);
if (c == SPACE || i==0) {
index = (i==0)?0:i+1;
temp.append(string.substring(index, last));
if (index!=0) temp.append(c);
builder.append(temp);
temp.delete(0, temp.length());
last = i;
}
}
return builder.toString();
}
In Java in-place:
public static final String reverseWordsInPlace(String string) {
char[] chars = string.toCharArray();
int lengthI = 0;
int lastI = 0;
int lengthJ = 0;
int lastJ = chars.length-1;
int i = 0;
char iChar = 0;
char jChar = 0;
while (i<chars.length && i<=lastJ) {
iChar = chars[i];
if (iChar == SPACE) {
lengthI = i-lastI;
for (int j=lastJ; j>=i; j--) {
jChar = chars[j];
if (jChar == SPACE) {
lengthJ = lastJ-j;
swapWords(lastI, i-1, j+1, lastJ, chars);
lastJ = lastJ-lengthI-1;
break;
}
}
lastI = lastI+lengthJ+1;
i = lastI;
} else {
i++;
}
}
return String.valueOf(chars);
}
private static final void swapWords(int startA, int endA, int startB, int endB, char[] array) {
int lengthA = endA-startA+1;
int lengthB = endB-startB+1;
int length = lengthA;
if (lengthA>lengthB) length = lengthB;
int indexA = 0;
int indexB = 0;
char c = 0;
for (int i=0; i<length; i++) {
indexA = startA+i;
indexB = startB+i;
c = array[indexB];
array[indexB] = array[indexA];
array[indexA] = c;
}
if (lengthB>lengthA) {
length = lengthB-lengthA;
int end = 0;
for (int i=0; i<length; i++) {
end = endB-((length-1)-i);
c = array[end];
shiftRight(endA+i,end,array);
array[endA+1+i] = c;
}
} else if (lengthA>lengthB) {
length = lengthA-lengthB;
for (int i=0; i<length; i++) {
c = array[endA];
shiftLeft(endA,endB,array);
array[endB+i] = c;
}
}
}
private static final void shiftRight(int start, int end, char[] array) {
for (int i=end; i>start; i--) {
array[i] = array[i-1];
}
}
private static final void shiftLeft(int start, int end, char[] array) {
for (int i=start; i<end; i++) {
array[i] = array[i+1];
}
}
Here is a C implementation that is doing the word reversing inlace, and it has O(n) complexity.
char* reverse(char *str, char wordend=0)
{
char c;
size_t len = 0;
if (wordend==0) {
len = strlen(str);
}
else {
for(size_t i=0;str[i]!=wordend && str[i]!=0;i++)
len = i+1;
}
for(size_t i=0;i<len/2;i++) {
c = str[i];
str[i] = str[len-i-1];
str[len-i-1] = c;
}
return str;
}
char* inplace_reverse_words(char *w)
{
reverse(w); // reverse all letters first
bool is_word_start = (w[0]!=0x20);
for(size_t i=0;i<strlen(w);i++){
if(w[i]!=0x20 && is_word_start) {
reverse(&w[i], 0x20); // reverse one word only
is_word_start = false;
}
if (!is_word_start && w[i]==0x20) // found new word
is_word_start = true;
}
return w;
}
c# solution to reverse words in a sentence
using System;
class helloworld {
public void ReverseString(String[] words) {
int end = words.Length-1;
for (int start = 0; start < end; start++) {
String tempc;
if (start < end ) {
tempc = words[start];
words[start] = words[end];
words[end--] = tempc;
}
}
foreach (String s1 in words) {
Console.Write("{0} ",s1);
}
}
}
class reverse {
static void Main() {
string s= "beauty lies in the heart of the peaople";
String[] sent_char=s.Split(' ');
helloworld h1 = new helloworld();
h1.ReverseString(sent_char);
}
}
output:
peaople the of heart the in lies beauty Press any key to continue . . .
Better version
Check my blog http://bamaracoulibaly.blogspot.co.uk/2012/04/19-reverse-order-of-words-in-text.html
public string reverseTheWords(string description)
{
if(!(string.IsNullOrEmpty(description)) && (description.IndexOf(" ") > 1))
{
string[] words= description.Split(' ');
Array.Reverse(words);
foreach (string word in words)
{
string phrase = string.Join(" ", words);
Console.WriteLine(phrase);
}
return phrase;
}
return description;
}
public class manip{
public static char[] rev(char[] a,int left,int right) {
char temp;
for (int i=0;i<(right - left)/2;i++) {
temp = a[i + left];
a[i + left] = a[right -i -1];
a[right -i -1] = temp;
}
return a;
}
public static void main(String[] args) throws IOException {
String s= "i think this works";
char[] str = s.toCharArray();
int i=0;
rev(str,i,s.length());
int j=0;
while(j < str.length) {
if (str[j] != ' ' && j != str.length -1) {
j++;
} else
{
if (j == (str.length -1)) {
j++;
}
rev(str,i,j);
i=j+1;
j=i;
}
}
System.out.println(str);
}
I know there are several correct answers. Here is the one in C that I came up with.
This is an implementation of the excepted answer. Time complexity is O(n) and no extra string is used.
#include<stdio.h>
char * strRev(char *str, char tok)
{
int len = 0, i;
char *temp = str;
char swap;
while(*temp != tok && *temp != '\0') {
len++; temp++;
}
len--;
for(i = 0; i < len/2; i++) {
swap = str[i];
str[i] = str[len - i];
str[len - i] = swap;
}
// Return pointer to the next token.
return str + len + 1;
}
int main(void)
{
char a[] = "Reverse this string.";
char *temp = a;
if (a == NULL)
return -1;
// Reverse whole string character by character.
strRev(a, '\0');
// Reverse every word in the string again.
while(1) {
temp = strRev(temp, ' ');
if (*temp == '\0')
break;
temp++;
}
printf("Reversed string: %s\n", a);
return 0;
}