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C++实现LeetCode(186.翻转字符串中的单词之二)

发布时间:2021-8-5 10:17

[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: 

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);
        }
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/186

类似题目:

Reverse Words in a String III

Reverse Words in a String

Rotate Array

参考资料:

https://leetcode.com/problems/reverse-words-in-a-string-ii/

https://leetcode.com/problems/reverse-words-in-a-string-ii/discuss/53851/Six-lines-solution-in-C%2B%2B

https://leetcode.com/problems/reverse-words-in-a-string-ii/discuss/53775/My-Java-solution-with-explanation

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