151. Reverse Words in a String

Aurora20052024-07-18 13:12

Given an input string s, reverse the order of the words.

A word is defined as a sequence of non-space characters. The words in s will be separated by at least one space.

Return a string of the words in reverse order concatenated by a single space.

Note that s may contain leading or trailing spaces or multiple spaces between two words. The returned string should only have a single space separating the words. Do not include any extra spaces.

Example 1:

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Input: s = "the sky is blue"
Output: "blue is sky the"

Example 2:

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Input: s = "  hello world  "
Output: "world hello"
Explanation: Your reversed string should not contain leading or trailing spaces.

Example 3:

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Input: s = "a good   example"
Output: "example good a"
Explanation: You need to reduce multiple spaces between two words to a single space in the reversed string.

Constraints:

1 <= s.length <= 104
s contains English letters (upper-case and lower-case), digits, and spaces ' '.
There is at least one word in s.

Follow-up: If the string data type is mutable in your language, can you solve it in-place with O(1) extra space?

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class Solution {
public:
    void reverse(string &s,int start,int end){
        for(int i=start,j=end-1;i<j;i++,j--){
            swap(s[i],s[j]);
        }
    }
    void removeExtraSpace(string &s){
        int slow=0;
        for(int fast=0;fast<s.size();fast++){
            if(s[fast]!=' '){
                if(slow!=0)s[slow++]=' ';
                while(fast<s.size() && s[fast]!=' ')s[slow++]=s[fast++];
            }
        }
        s.resize(slow);
    }
    string reverseWords(string s) {
        removeExtraSpace(s);
        reverse(s,0,s.size());
        int start=0;
        for(int i=0;i<=s.size();i++){
            if(i==s.size() || s[i]==' '){
                reverse(s,start,i);
                start=i+1;
            }
        }
        return s;
    }
};

注意：

1.需要删除多余空格，但是每个单词之间又需要一个空格

2.总体思路：先把多余空格删除，再把整个字符串反转，再依次把每一个单词反转

3.reverse函数很简单，不过多赘述

4.removeExtraSpace函数：使用快慢指针，重点是当fast所在位置不为空格且slow>0就需要手动添加一个空格。