leetcode 541. Reverse String II

原题：

Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and left the other as original.

Example:

Input: s = "abcdefg", k = 2Output: "bacdfeg"

Restrictions:

1. The string consists of lower English letters only.
2. Length of the given string and k will in the range [1, 10000]

每2k个元素反转前k个，不够k个的反转前k个。

代码如下：

char* reverseStr(char* s, int k) {    int len = strlen(s);    char temp;    char* org=s;    while(len!=0)    {        if(len>k)        {            for(int n=0;n<k-n-1;n++)            {                temp=*(s+n);                *(s+n)=*(s+k-n-1);                *(s+k-n-1)=temp;            }            if(len>2*k)            {                s=s+2*k;                len=strlen(s);            }            else            {                len=0;            }        }        else         {            for(int n=0;n<len-n-1;n++)            {                temp=*(s+n);                *(s+n)=*(s+len-n-1);                *(s+len-n-1)=temp;            }            len=0;        }    }    return org;}

就是简单的顺序2k扫描