Leetcode: Reverse Words in a String

Given an input string, reverse the string word by word.

For example,

Given s = "

the sky is blue

",

return "

blue is sky the

".

click to show clarification.

Clarification:

What constitutes a word?

A sequence of non-space characters constitutes a word.
Could the input string contain leading or trailing spaces?

Yes. However, your reversed string should not contain leading or trailing spaces.
How about multiple spaces between two words?

Reduce them to a single space in the reversed string.
最新加上的题，原理比较简单，两次翻转，但实现起来需要注意边界用例。因为可能多个空格，所以先分割单词会比较好。

class Solution {
public:
void reverseWords(string &s) {
int size = s.size();
int start = 0, end = 0;

string word;
int i = 0;
while (i < size) {
if (s[i] == ' ' && start < i) {
// Abstract word and reverse
word = s.substr(start, i - start);
reverseWord(word);
for (int idx = 0; idx < word.size(); ++idx) {
s[end++] = word[idx];
}
s[end++] = ' ';
}
while (s[i] == ' ' && i + 1 < size && s[i+1] == ' ') {
++i;
}
if (s[i] == ' ') {
start = i + 1;
}
++i;
}
if (start < size) {
word = s.substr(start, i - start);
reverseWord(word);
for (int idx = 0; idx < word.size(); ++idx) {
s[end++] = word[idx];
}
}

// Eliminate the trailing space
if (s[end-1] == ' ') {
--end;
}

s = s.substr(0, end);
reverseWord(s);
}

void reverseWord(string &word) {
char tmp;
for (int i = 0, j = word.size() - 1; i < j; ++i, --j) {
tmp = word[i];
word[i] = word[j];
word[j] = tmp;
}
}
};

=======================第二次=========================

<pre name="code" class="cpp">class Solution {
public:
void reverseWords(string &s) {
int cur = 0;
int word_start = 0;
int i = 0;
while (i < s.size()) {
if (s[i] == ' ' && (cur == 0 || s[i-1] == ' ')) {
++i;
}
else {
if (s[i] == ' ') {
reverse(s, word_start, cur - 1);
word_start = cur + 1;
}
s[cur++] = s[i++];
}
}

--cur;
if (cur > 0) {
if (s[cur] == ' ') {
--cur;
}
else {
reverse(s, word_start, cur);
}
}

reverse(s, 0, cur);
s = s.substr(0, cur + 1);
}

void reverse(string &s, int start, int end) {
char tmp;
for (; start < end; ++start, --end) {
tmp = s[start];
s[start] = s[end];
s[end] = tmp;
}
}
};