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541. Reverse String II

541. Reverse String II

作者: matrxyz | 来源:发表于2018-01-13 16:06 被阅读0次

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"

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

Solution:遍历

思路:
Time Complexity: O(N) Space Complexity: O(1)

Solution Code:

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 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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