LeetCode 557. Reverse Words in a String III

Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.

Example 1:

Input: "Let's take LeetCode contest"
Output: "s'teL ekat edoCteeL tsetnoc"

Note: In the string, each word is separated by single space and there will not be any extra space in the string.

题目的要求很简单，要求翻转每个单词，但是保持空格以及单词的相对顺序不变，最开始我想的是用一个栈，遇见空格把栈中的单词弹出，最后也确实AC了，代码如下：

class Solution {
public:
    string reverseWords(string s) {
        stack<char> sta;
        //s += ' ';
        string res = "";
        for (int i = 0; i <= s.length(); i++)
        {
           if (s[i] != ' ' && i != (s.length()))
           //if (s[i] != ' ')
            {
                sta.push(s[i]); 
                //cout<<sta.top()<<endl;
            }
            else
            {
               while (!sta.empty())
               {
                 res += sta.top();
                 //std::cout<<sta.top();
                 sta.pop();
               }
               if (s[i] == ' ')
                 res += ' ';
            }
        }
        cout<<endl;
        return res;
    }
};

当然也可以使用reverse函数，代码如下：

class Solution {
public:
    string reverseWords(string s) {
        int front = 0;
        for (int i = 0; i <= s.length(); i++)
        {
            if (s[i] == ' ' || i == s.length())
            {
                reverse(&s[front], &s[i]);
            front = i + 1;
            }
                
        }
        return s;
    }
};

或者是使用两个指针，分别指向字符串的 头和尾部：

class Solution {
public:
    string reverseWords(string s) {
        int start = 0, end = 0, n = s.size();
        while (start < n && end < n) {
            while (end < n && s[end] != ' ') ++end;
            for (int i = start, j = end - 1; i < j; ++i, --j) {
                swap(s[i], s[j]);
            }
            start = ++end;
        }
        return s;
    }
};