• 一、题目
  • 二、解题思路
  • 三、解题代码

    一、题目

    Given a string and an offset, rotate string by offset. (rotate from left to right)

    Example

    Given "abcdefg".

    offset=0 => “abcdefg”
    offset=1 => “gabcdef”
    offset=2 => “fgabcde”
    offset=3 => “efgabcd”

    Challenge

    Rotate in-place with O(1) extra memory.

    给定一个字符串和一个偏移量,根据偏移量旋转字符串(从左向右旋转)

    二、解题思路

    常见的翻转法应用题,仔细观察规律可知翻转的分割点在从数组末尾数起的offset位置。先翻转前半部分,随后翻转后半部分,最后整体翻转。

    三、解题代码

    1. public class Solution {
    2. /*
    3. * param A: A string
    4. * param offset: Rotate string with offset.
    5. * return: Rotated string.
    6. */
    7. public char[] rotateString(char[] A, int offset) {
    8. if (A == null || A.length == 0) {
    9. return A;
    10. }
    11. int len = A.length;
    12. offset %= len;
    13. reverse(A, 0, len - offset - 1);
    14. reverse(A, len - offset, len - 1);
    15. reverse(A, 0, len - 1);
    16. return A;
    17. }
    18. private void reverse(char[] str, int start, int end) {
    19. while (start < end) {
    20. char temp = str[start];
    21. str[start] = str[end];
    22. str[end] = temp;
    23. start++;
    24. end--;
    25. }
    26. }
    27. }