C++实现LeetCode(186.翻转字符串中的单词之二)

来源：jb51　　时间：2021/8/4 17:55:50　　对本文有异议

[LeetCode] 186. Reverse Words in a String II 翻转字符串中的单词之二

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

Example:

Input: ["t","h","e"," ","s","k","y"," ","i","s"," ","b","l","u","e"]
Output: ["b","l","u","e"," ","i","s"," ","s","k","y"," ","t","h","e"]

Note:

A word is defined as a sequence of non-space characters.
The input string does not contain leading or trailing spaces.
The words are always separated by a single space.

Follow up: Could you do it in-place without allocating extra space?

这道题让我们翻转一个字符串中的单词，跟之前那题 Reverse Words in a String 没有区别，由于之前那道题就是用 in-place 的方法做的，而这道题反而更简化了题目，因为不考虑首尾空格了和单词之间的多空格了，方法还是很简单，先把每个单词翻转一遍，再把整个字符串翻转一遍，或者也可以调换个顺序，先翻转整个字符串，再翻转每个单词，参见代码如下：

解法一：

class Solution {
public:
    void reverseWords(vector<char>& str) {
        int left = 0, n = str.size();
        for (int i = 0; i <= n; ++i) {
            if (i == n || str[i] == ' ') {
                reverse(str, left, i - 1);
                left = i + 1;
            }
        }
        reverse(str, 0, n - 1);
    }
    void reverse(vector<char>& str, int left, int right) {
        while (left < right) {
            char t = str[left];
            str[left] = str[right];
            str[right] = t;
            ++left; --right;
        }
    }
};

我们也可以使用 C++ STL 中自带的 reverse 函数来做，先把整个字符串翻转一下，然后再来扫描每个字符，用两个指针，一个指向开头，另一个开始遍历，遇到空格停止，这样两个指针之间就确定了一个单词的范围，直接调用 reverse 函数翻转，然后移动头指针到下一个位置，在用另一个指针继续扫描，重复上述步骤即可，参见代码如下：

解法二：

class Solution {
public:
    void reverseWords(vector<char>& str) {
        reverse(str.begin(), str.end());
        for (int i = 0, j = 0; i < str.size(); i = j + 1) {
            for (j = i; j < str.size(); ++j) {
                if (str[j] == ' ') break;
            }
            reverse(str.begin() + i, str.begin() + j);
        }
    }
};