Leetcode 151.反转字符串中的单词

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此题是非常经典的字符串的颠倒问题，但这个更复杂一些，但也不其本质，我此次写的方式是用双指针问题完成的，虽然算不上什么好方法，但如果各位看官觉得满意的话，请各位给我个点个免费的赞吧，谢谢了^_

1.题目要求如图所示:

2.接下来是做题的步骤:

我们先把字符串的颠倒函数写好,如图所示:

void reverse(char* left,char* right)
{
    while(left <= right)
    {
        char str = *left;
        *left = *right;
        *right = str;
        left++;
        right--;
    }
}

(1).我们先把整个字符串都颠倒过来,如图所示:

char* left = s;
    char* right = s + strlen(s) - 1;
    //先进行字符串逆置
    reverse(left,right);

(2).然后再让各个单词颠倒过来,如图所示:

char* start = s;
    while(*start != '\0')
    {
        while(*start == ' '&&*start != '\0')
        {
            start++;
        }
        char* end = start;
        while(*end != ' '&&*end != '\0')
        {
            end++;
        }
        reverse(start,end-1);
        if(*end != '\0')
        {
            end++;
        }
        start = end;
    }

(3)此时各个单词颠倒过来后,我们就先把字符串首尾的括号去掉，我是用malloc函数辅助我这一想法的^_,如图所示:

left = s;
    right = s + strlen(s) - 1;
    //头尾括号
    while (*left == ' ')
        left++;
    while(*right == ' ')
        right--;
    char* string = (char*)malloc(sizeof(char) * (strlen(s) + 1));
    int j = 0;
    while(left <= right&&*left != '\0')
    {
        *(string + j) = *left;
        j++;
        left++;
    }
    *(string + j) = '\0';

(4).然后我们再去掉单词之间的括号，如图所示:

int s_size = 0;
    int i = 0;
    while(i < j)
    {
        if(string[i] == ' '&&string[i] != '\0')
        {
            s[s_size] = string[i];
            s_size++;
            i++;
            if(i >= j)
            break;
            while(string[i] == ' ')
            {
                i++;
                if(i >= j)
                break;
            }
        }else{
            s[s_size] = string[i];
            s_size++;
            i++;
            if(i >= j)
            break;
        }
    }
    s[s_size] = '\0';

(5).全部代码如下图所示:

//字符串逆置函数
void reverse(char* left,char* right)
{
    while(left <= right)
    {
        char str = *left;
        *left = *right;
        *right = str;
        left++;
        right--;
    }
}
char* reverseWords(char* s) {
    char* ret = s;
    char* left = s;
    char* right = s + strlen(s) - 1;
    //先进行字符串逆置
    reverse(left,right);
    //然后再让单词进行逆置
    char* start = s;
    while(*start != '\0')
    {
        while(*start == ' '&&*start != '\0')
        {
            start++;
        }
        char* end = start;
        while(*end != ' '&&*end != '\0')
        {
            end++;
        }
        reverse(start,end-1);
        if(*end != '\0')
        {
            end++;
        }
        start = end;
    }
    left = s;
    right = s + strlen(s) - 1;
    //去掉头尾括号
    while (*left == ' ')
        left++;
    while(*right == ' ')
        right--;
    char* string = (char*)malloc(sizeof(char) * (strlen(s) + 1));
    int j = 0;
    while(left <= right&&*left != '\0')
    {
        *(string + j) = *left;
        j++;
        left++;
    }
    *(string + j) = '\0';
    //去掉单词之间的括号
    int s_size = 0;
    int i = 0;
    while(i < j)
    {
        if(string[i] == ' '&&string[i] != '\0')
        {
            s[s_size] = string[i];
            s_size++;
            i++;
            if(i >= j)
            break;
            while(string[i] == ' ')
            {
                i++;
                if(i >= j)
                break;
            }
        }else{
            s[s_size] = string[i];
            s_size++;
            i++;
            if(i >= j)
            break;
        }
    }
    s[s_size] = '\0';
    return ret;
}

好了，各位看官，这就是我的代码，各位看官如果觉得满意的话就给个免费的赞吧，谢谢各位了。

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