[Leetcode] 557. Reverse Words in a String III

最新推荐文章于 2024-01-04 11:20:32 发布

原创最新推荐文章于 2024-01-04 11:20:32 发布 · 279 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #Reverse

LeetCode 专栏收录该内容

80 篇文章

订阅专栏

博客围绕字符串处理展开，要求反转句子中每个单词的字符顺序，同时保留空格和单词初始顺序。给出示例，输入‘Let’s take LeetCode contest’，输出‘s’teL ekat edoCteeL tsetnoc’，并分析该题比相关题目更简单，只需对每个单词执行反转。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

分析：
其实这一题比151 Reverse Words in a String 更简单，对一个字符串进行反转，没有多余的空格。
只需要对字符串中的每一个单词执行反转即可。

void _reverse(char *pStart, char *pEnd)
{
    if(pStart >= pEnd)
        return;
    char c ;
    while(pStart < pEnd){
        c = *pStart;
        *pStart = *pEnd;
        *pEnd = c;
        pStart++;
        pEnd--;
    }
}
char* reverseWords(char* s) {
   int length = strlen(s);
   if(length <= 1)
       return s;
   char *pStart = s;
   char *pEnd = s+length -1;
   char *p ;
    while(pStart <pEnd){
        p=pStart;
        while(*p!= ' ' && *p!='\0') p++;
        _reverse(pStart, p-1);
        pStart = ++p;
    }
    return s;
}