Rotate Array

问题描述

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].

Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.

Hint:
Could you do it in-place with O(1) extra space?

算法

代码一

 1 public void rotate(int[] nums, int k) {
 2         int n = nums.length;
 3         int temp;
 4         k = k % n;
 5         for (int i = 0; i < n / 2; i++) {
 6             temp = nums[i];
 7             nums[i] = nums[n - 1 - i];
 8             nums[n - 1 - i] = temp;
 9         }
10         for (int i = 0; i < k / 2; i++) {
11             temp = nums[i];
12             nums[i] = nums[k - 1 - i];
13             nums[k - 1 - i] = temp;
14         }
15         for (int i = 0; i < (n - k) / 2; i++) {
16             temp = nums[k + i];
17             nums[k + i] = nums[n - 1 - i];
18             nums[n - 1 - i] = temp;
19         }
20     }

注意k可能会比n大，所以将k模n。

代码二

 1 public void rotate1(int[] nums,int k){
 2         int n=nums.length;
 3         k=k%n;
 4         reverse(nums,0,n-1);
 5         reverse(nums,0,k-1);
 6         reverse(nums,k,n-1);
 7     }
 8     private void reverse(int[] nums,int s,int e){
 9         int temp;
10         while(s<e){
11             temp=nums[s];
12             nums[s]=nums[e];
13             nums[e]=temp;
14             s++;
15             e--;
16         }
17     }
18     private void reverse1(int[] nums,int m,int n){
19         while(m<n){
20             nums[m]^=nums[n];
21             nums[n]^=nums[m];
22             nums[m]^=nums[n];
23             m++;
24             n--;
25         }
26     }

代码二的思想和代码一的思想一样，优点是把三次反转写成一个函数。另外两个值交换可以用异或运算。

代码三

 1 //https://leetcode.com/discuss/28501/my-c-solution-o-n-time-%26%26-o-1-space
 2     public void rotate2(int[] nums, int k) {
 3         int n = nums.length;
 4         k = k % n;
 5         int i = 0;
 6         int nowIndex = 0, tmp1, tmp2 = nums[0];
 7         for (int j = 0; j < n; j++) {
 8             tmp1 = tmp2;
 9             nowIndex = (nowIndex + k) % n;
10             tmp2 = nums[nowIndex];
11             nums[nowIndex] = tmp1;
12             if (nowIndex == i && i < n - 1) {
13                 i++;
14                 nowIndex = i;
15                 tmp2 = nums[nowIndex];
16             }
17 
18         }
19     }