541. Reverse String II

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541. Reverse String II

题目描述：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"
```
题目含义：给定一个字符串和一个整数k，需要将每2k个字符中的前K个字符进行翻转，如果不足2k个字符但是大于k个字符，则将前k个字符翻转，如果少于K个字符则保持原样。
思路：见代码

代码

class Solution {public String reverseStr(String s, int k) {      char[] arr = s.toCharArray();      int i = 0;      int n = s.length();      while (i < s.length()) {          int j = Math.min(i + k - 1, n - 1);          swap(arr, i, j);          i += 2 * k;      }      return String.valueOf(arr);  }  private void swap(char[] arr, int l, int r) {      while (l < r) {          char temp = arr[l];          arr[l++] = arr[r];          arr[r--] = temp;      }  }}

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