【LeetCode】Reverse Words in a String 反转字符串中的单词

一年没有管理博客园了，说来实在惭愧。。

最近开始刷LeetCode，之前没刷过，说来也实在惭愧。。。

刚开始按 AC Rates 从简单到难刷，觉得略无聊，就决定按 Add Date 刷，以后也可能看心情随便选题…(⊙o⊙)…今天做了14年 Add 的三个题，其中 Maximum Product Subarray 着实把我折腾了好一会儿，所以想想还是跟大家分享一下我的解法，也或许有更好的方法，期待大家的分享。就把三个题按 Add Date 的先后顺序分享一下吧。

Add Date 2014-03-05

Reverse Words in a String

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

For example,
Given s = "

the sky is blue

",
return "

blue is sky the

".

Clarification:

What constitutes a word?
A sequence of non-space characters constitutes a word.

Could the input string contain leading or trailing spaces?
Yes. However, your reversed string should not contain leading or trailing spaces.

How about multiple spaces between two words?
Reduce them to a single space in the reversed string.

单纯把 "

the sky is blue

" reverse 成 "

blue is sky the

" 是件很容易的事情，也是比较经典的题目，不过本题说：

1. s 中的开头和结尾有可能有空格，reverse 后的 string 中要去掉；

2. 连续多个空格要变成一个空格。

我的 code 略折腾，用了两个函数：

1. reverseWord 实现最基本的 reverse，首先整个 reverse，变为“eulb si yks eht”，然后每个 word reverse，遍历两边即可。

2. removeSpace 实现检查和删除空格。遍历一遍即可。

复杂度O(n).

附 code，仅供参考。

class Solution {
public:
void reverseWord(string &s) {                       //反转字符串
string::iterator pre = s.begin();
string::iterator post = s.end()-1;
char tmp;
while(pre < post) {                             //反转整个字符串
tmp = *pre;
*pre = *post;
*post = tmp;
++pre;
--post;
}
pre = s.begin();
post = pre;
while(post != s.end()) {                        //反转每个单词
while(pre != s.end() && *pre == ' ') {
++pre;
}
if(pre == s.end())
return;
post = pre;
while(post != s.end() && *post != ' ') {
++post;
}
--post;
if(post != s.end() && pre < post) {
string::iterator p1 = pre;
string::iterator p2 = post;
while(p1 < p2) {
tmp = *p1;
*p1 = *p2;
*p2 = tmp;
++p1;
--p2;
}
}
++post;
pre = post;
}
}

void removeSpace(string &s) {                       //检查和删除空格
string::iterator pre = s.begin();
string::iterator post = pre;
while(pre != s.end() && *pre == ' ') {          //删除开头的空格
s.erase(s.begin());
pre = s.begin();
}
if(pre == s.end())
return;
post = pre;
while(post != s.end()) {                        //把连续多个空格变为一个
while(pre != s.end() && *pre != ' ')
++pre;
if(pre == s.end())
return;
post = pre+1;
while(post != s.end() && *post == ' ') {
s.erase(post);
post = pre+1;
}
++pre;
}
if(!s.empty()) {                                //如果最后有空格则删除
pre = s.end()-1;
if(*pre == ' ')
s.erase(pre);
}
}

void reverseWords(string &s) {
reverseWord(s);
removeSpace(s);
}
};

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