算法1

class Solution {
public:
    string reverseWords(string s) {
        if (!s.size()) return ""; // 特判
        vector<string> words;     // 存单词
        for (int i = 0, j = 0; i < s.size(); i = ++ j) // i为快指针，j为慢指针，i = ++ j是为了跳出空格，找到下一个单词
        {
            while (s[j] != ' ' && j < s.size()) j ++ ; // 找到单词结尾
            words.push_back(s.substr(i, j - i));  // 将单词存入vector
        }
        reverse(words.begin(), words.end());   // 反转数组
        string res;  // 答案
        for (int i = 0; i < words.size(); i ++ )
        {
            if (i != words.size() - 1) res += words[i] + ' ';
            else res += words[words.size() - 1];
        }
        return res;
    }
};

算法2

class Solution {
public:
    string reverseWords(string s) {
        reverse(s.begin(), s.end());
        s.push_back(' ');
        for (int i = 0; i < s.size(); )
        {
            int l = i, r = i;
            while (s[r] != ' ') r ++ ;
            i = r + 1;
            while (l < r)
            {
                swap(s[l], s[r]);
                l ++ , r -- ;
            }
        }
        return s.substr(1);
    }
};