标签：数组 within rac example bsp 代码 一个 front 遍历
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.
题意：把输入的字符串单词的顺序不变，单词中字母顺序逆置，测试中所给的字符串中不会有两个相连的空格。
基本思路：
1. 查单词，把从开始到空格、空格之间和空格到结束之间的单词分别保存在一个字符串数组中，当结束时，把字符串数组中的每个单词倒置，然后按顺序输出中间插入空格。
这种方法虽然条理清晰，但是用了大量的额外空间，而且相当于要遍历三次，所以不推荐使用。
2. 下面这种方式没有使用额外空间就对每个单词进行头尾交换。代码如下
1 char* reverseWords(char* s) { 2 char *beg = s; 3 char *end = s; 4 char temp = 0; 5 while(*beg != ‘\0‘) 6 { 7 if(*beg == ‘ ‘){ 8 ++beg; 9 ++end; 10 }else if(*end == ‘\0‘ || *end == ‘ ‘) 11 { 12 char *ps = beg, *pe = end-1; 13 while(ps <pe){ 14 temp = *ps; 15 *ps = *pe; 16 *pe = temp; 17 ps++; 18 pe--; 19 } 20 beg = end; 21 }else 22 end++; 23 } 24 return s; 25 }
写的简洁一点就是
1 class Solution { 2 public: 3 string reverseWords(string s) { 4 size_t front = 0; 5 for(int i = 0; i <= s.length(); ++i) 6 if(i == s.length() || s[i] == 0x20){ 7 reverse(&s[front], &s[i]); 8 front = i + 1; 9 } 10 return s; 11 } 12 };
使用c++的reverse函数而不自定义反转函数，关于reverse函数：http://www.cplusplus.com/reference/algorithm/reverse/
LeetCode解题思路：557. Reverse Words in a String III
标签：数组 within rac example bsp 代码 一个 front 遍历
原文地址：http://www.cnblogs.com/hellomotty/p/7392133.html
