Leetcode: Rotate Array

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.

[show hint]

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

Naive想法就是保存一个原数组的拷贝，然后把原数组分成前len-k个元素和后k个元素两部分，把后k个元素放到前len-k个元素前面去。这样做需要O(N)空间

in-place做法是： 

(1) reverse the array;

(2) reverse the first k elements;

(3) reverse the last n-k elements.

The first step moves the first n-k elements to the end, and moves the last k elements to the front. The next two steps put elements in the right order.

public class Solution {
    public void rotate(int[] nums, int k) {
        int len = nums.length;
        k %= len;
        reverse(nums, 0, len-1);
        reverse(nums, 0, k-1);
        reverse(nums, k, len-1);
    }
    
    public void reverse(int[] nums, int l, int r) {
        while (l <= r) {
            int temp = nums[l];
            nums[l] = nums[r];
            nums[r] = temp;
            l++;
            r--;
        }
    }
}

需要注意的是第4行，右移偏移量k可能比数组长度len要大，所以要先 k%=len；