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

"

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.

使用stack结构存储每一个单词，之后弹出就是逆序排列了。

class Solution {
public:
void reverseWords(string &s) {
stack<string> reverse;
int index=0;
while(s[index]!='\0'){
string temp="";
while(s[index]!='\0'&&s[index]==' '){
index++;
}
while(s[index]!='\0'&&s[index]!=' '){
temp+=s[index];
index++;
}
reverse.push(temp);
}
string result="";
while(!reverse.empty()){
/*  string temp=reverse.top();
reverse.pop();
result+=temp;
if(temp!=""&&!reverse.empty()){  //在这里需要注意的是，temp何时为空。只有当末端有一个或多个空格的时候，才会为空。这时候逆转过来就不应该加入字符串，并且最后一个元素后面也不应该加入空格
result+=" ";

<span style="white-space:pre">		</span>}<span style="font-family: Arial, Helvetica, sans-serif;">*/</span>

 <span style="white-space:pre">	</span>    string temp=reverse.top(); /*换一种方式，在单词前面加空格*/
            reverse.pop();
            if(!result.empty()){
                result+=" ";
            }
            result+=temp;

}
s=result;
}
};

还有一种方法：

1 从后往前遍历string

2 保存单词，然后需要逆转，在保存到结果中

注意题目中的clarification，处理好其中的空格。

利用了两个额外string保存中间结果。空间复杂度为O(n).

class Solution {
public:
void reverseWords(string &s)
{
string re;
for(int i=s.length()-1;i>=0;){
while(i>=0&&s[i]==' ') i--;
if(i<0) break;
if(!re.empty()){
re.push_back(' ');
}
string temp;
while(i>=0&&s[i]!=' ') temp.push_back(s[i--]);
reverse(temp.begin(),temp.end());
re.append(temp);
}
s=re;
}
};