151.Reverse Words in a String[medium]

最新推荐文章于 2025-07-29 14:24:32 发布

鹅鹅鹅下个东西好难

最新推荐文章于 2025-07-29 14:24:32 发布

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本文介绍了一种将输入字符串中的单词逆序输出的算法。首先给出了一种简单但空间复杂度较高的方法，即将字符串分割为单词列表并逆序，再重新组合。随后提出了一种空间复杂度为O(1)的C语言实现方案，该方案通过两次逆序操作来达到目的。

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

For example,
Given s = "the sky is blue",
return "blue is sky the".

Update (2015-02-12):
For C programmers: Try to solve it in-place in O(1) space.

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.

最基本的解决方案，非原地翻转，用python和c可以很容易解决，但空间复杂度较大，思路是先split成数组，对数组做翻转再还原

class Solution:  
    # @param s, a string  
    # @return a string  
    def reverseWords(self, s):  
        L = s.split()#字符串以单词分隔成列表  
        L.reverse()#反转列表中的单词  
        s1 = ' '.join(L)#以空格分隔列表中反转后的单词，返回字符串  
        return s1

但是基于原地翻转空间复杂度O（1）的方法就没法用python了，使用c解决, 思路是先给每个单词词内进行翻转，然后对整个句子以字母为最小单位翻转，最后去除掉多余的空格。

void reverseWords(char *s) {
    
    int i = 0;
    int start=0;
    int end=0;
    while (s[i]){

        if (i == 0 || (s[i]!= ' ' && s[i-1]==' ') ) {
            start = i;
        }
        if (!s[i] || (s[i]!=' ' && s[i+1] == ' ') ){
            end = i;
            reverse_substring(s,start,end);

        }
        printf("i = %d, start = %d, end = %d ,%s\n",i,start,end,s);
        i++;

    }

    int last = i-1;
    reverse_substring(s,start,last);
    reverse_substring(s,0,last); 
    
        //check '   '
    i = 0;
    int lastLetter = -1;
    while (s[i]){
        if (s[i] != ' ') {
            lastLetter  = i;
        }
        i++;
    }

    s[lastLetter+1] = '\0';
}


void reverse_substring(char *s,int start,int end){
    for(int i=0;i<=(end-start)/2;i++)
    {
        char tmp = s[start+i];
        s[start+i] = s[end-i];
        s[end - i] = tmp;
    }
}