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541. Reverse String II

时间:2019-04-09 18:02:48      阅读:112      评论:0      收藏:0      [点我收藏+]

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 = 2 Output: "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个字符里,翻转前k个
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); //i的值确定后,j的值为i + k - 1,要判断是否超出数组(和n - 1比较)
            swap(arr, i, j);
            i += 2 * k;
        }
        return new String(arr);
       //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;
        }
    }
}

541. Reverse String II

原文:https://www.cnblogs.com/MarkLeeBYR/p/10678577.html

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