leetcode summary-section II

151 Reverse Words in a String

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

细节处理。

186 Reverse Words in a String II

class Solution {
public:
    void reverseWords(string& s) {
        reverse(s, 0, s.length);
        for (int i=0, j=0; j<=s.length; j++) {
            if (j==s.length || s[j]==' ') {
                reverse(s, i, j);
                i =  j + 1;
            }
        }
    }
 
    void reverse(string& s, int begin, int end) {
        while (begin < end - 1) {
            swap(s[begin], s[end - 1]);
            begin++; end--;
        }
    }
};

 是上一题的简化版。先翻转整个string，再逐个单词翻转。其中，专门用一个下标 i 来指示单词的起始位置。

233 number of digit 1s