# 541.Reverse String II

**541.Reverse String II**

难度:Easy

> 给定一个字符串和一个整数 k，你需要对从字符串开头算起的每个 2k 个字符的前k个字符进行反转。如果剩余少于 k 个字符，则将剩余的所有全部反转。如果有小于 2k 但大于或等于 k 个字符，则反转前 k 个字符，并将剩余的字符保持原样。

```
示例:

输入: s = "abcdefg", k = 2
输出: "bacdfeg"
要求:

该字符串只包含小写的英文字母。
给定字符串的长度和 k 在[1, 10000]范围内。
```

可以直接用字符串长度除以k，得到需要进行反转的字符串数目。最后将剩下的字符串加到结果中即可。

```
class Solution {
public:
    string reverseStr(string s, int k) {
        string res;
        int count=s.length()/k;
        for(int i=0;i<count;i++)
        {
            if(i%2) 
            {
                for(int j=0;j<k;j++)
                {
                    res+=s[i*k+j];
                }
            }
            else
            {
                for(int j=k-1;j>=0;j--)
                    res+=s[i*k+j];
            }
        }
        if(count%2)
        for(int i=count*k;i<s.length();i++)
            res+=s[i];
        else
            for(int i=s.length()-1;i>=count*k;i--)
            res+=s[i];
        return res;
    }
};
```

> 执行用时 :12ms, 在所有 C++ 提交中击败了69.60%的用户\
> 内存消耗 :10 MB, 在所有 C++ 提交中击败了21.26%的用户


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