LeetCode-541. Reverse String II (Java)

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]

----------------------------------------------------------------------------------------------------------------------------------------------------------------

题意

给定一个字符串和一个整数k，你需要反转2k长度字符串的前k个字符，如果剩余字符串长度小于k，则反转剩余字符串，

如果剩余字符串小于2k，大于或等于k，则反转剩余字符串前k个字符，其他字符顺序不变。

代码

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

我的代码多考虑了剩余字符串长度的判断，还有就是索引的处理有问题。

以后多注意。。。。