Rotate Array 解答

Question

Rotate an array of n elements to the right by k steps.

For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].

Solution 1 Bubble Rotate

Similar with bubble sort, time complexity O(k * n), space cost O(1)

 1 public class Solution {
 2     public void rotate(int[] nums, int k) {
 3         int length = nums.length;
 4         if (k >= length)
 5             k = k % length;
 6         if (k == 0)
 7             return;
 8         for (int i = 0; i < k; i++) {
 9             for (int j = length - 1; j > 0; j--) {
10                 int tmp = nums[j];
11                 nums[j] = nums[j - 1];
12                 nums[j - 1] = tmp;
13             }
14         }
15     }
16 }

Solution 2 Traverse

Key to this solution is to make a copy of original array. Time complexity O(n), space cost O(n).

Solution 3 Reversal

Assuming we are given {1,2,3,4,5,6} and order 2. The basic idea is:

1. Divide the array two parts: 1,2,3,4 and 5, 6

2. Rotate first part: 4,3,2,1,5,6

3. Rotate second part: 4,3,2,1,6,5

4. Rotate the whole array: 5,6,1,2,3,4

Time complexity O(n), space cost O(n)

 1 public class Solution {
 2     public void rotate(int[] nums, int k) {
 3         int length = nums.length;
 4         if (k >= length)
 5             k = k % length;
 6         if (k == 0)
 7             return;
 8         reverse(nums, 0, length - k - 1);
 9         reverse(nums, length - k, length - 1);
10         reverse(nums, 0, length - 1);
11     }
12     private void reverse(int[] nums, int start, int end) {
13         if (start == end || nums.length == 1)
14             return;
15         while (start < end) {
16             int tmp = nums[start];
17             nums[start] = nums[end];
18             nums[end] = tmp;
19             start++;
20             end--;
21         }
22     }
23 }