[力扣c语言实现]557. 反转字符串中的单词 III

557. 反转字符串中的单词 III

1. 题目描述

给定一个字符串,你需要反转字符串中每个单词的字符顺序,同时仍保留空格和单词的初始顺序。

示例:

输入:“Let’s take LeetCode contest”
输出:“s’teL ekat edoCteeL tsetnoc”

提示:

在字符串中,每个单词由单个空格分隔,并且字符串中不会有任何额外的空格。

2.代码如下

char * reverseWords(char * s){
    
    
    char *ret = NULL;
    char *p1 = NULL;
    char *p2 = NULL;
    char *tmp = NULL;
    int len = strlen(s);
	int len2 = 0;
    int incre = 0;    
     
    ret = (char *)malloc(len+1);
    memset(ret,0,len+1);

    if (len <= 0)
    {
    
    
        return ret;
    }

    p1 = s;
    while ((p2 = strchr(p1,' ')) != NULL)
    {
    
    
        tmp = p2;
        tmp -= 1;
        while (tmp != p1)
        {
    
    
            ret[incre++] = *tmp;
            tmp -= 1;
        }

        ret[incre++] = *p1;
        ret[incre++] = ' ';
        p1 = p2+1;
        p2 = NULL;
    }

	p2 = p1+strlen(p1)-1;
	
    while (p1 != p2)
	{
    
    
		ret[incre++] = *p2;
		p2 -=1;
	}
	
	ret[incre++] = *p1;
    return ret;
}

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转载自blog.csdn.net/dengwodaer/article/details/114447120