189. Rotate Array - Easy

Given an array, rotate the array to the right by k steps, where k is non-negative.

Example 1:

Input: [1,2,3,4,5,6,7] and k = 3
Output: [5,6,7,1,2,3,4]
Explanation:
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]

Example 2:

Input: [-1,-100,3,99] and k = 2
Output: [3,99,-1,-100]
Explanation: 
rotate 1 steps to the right: [99,-1,-100,3]
rotate 2 steps to the right: [3,99,-1,-100]

Note:

• Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
• Could you do it in-place with O(1) extra space?

 

in-place, O(1) extra space

先reverse前n-k个元素，再reverse后k个元素，再reverse整个数组

e.g. [1,2,3,4,5,6,7], k = 3 ：-> [4,3,2,1,5,6,7] -> [4,3,2,1,7,6,5] -> [5,4,7,1,2,3,4]

注意k有可能大于数组的长度，用mod找到最小的k

time: O(n), space: O(1)

class Solution {
    public void rotate(int[] nums, int k) {
        if(nums == null || nums.length == 0) return;
        int n = nums.length;
        k = k % n;
        reverse(nums, 0, n-1-k);
        reverse(nums, n-k, n-1);
        reverse(nums, 0, n-1);
    }
    
    private void reverse(int[] nums, int start, int end) {
        while(start <= end) {
            int tmp = nums[start];
            nums[start] = nums[end];
            nums[end] = tmp;
            start++;end--;
        }
    }
}

 

二刷：

class Solution {
    public void rotate(int[] nums, int k) {
        if(nums == null || nums.length == 0) {
            return;
        }
        int n = nums.length;
        k = k % n;
        reverse(nums, 0, n - 1);
        reverse(nums, 0, k - 1);
        reverse(nums, k, n - 1);
    }
    
    private void reverse(int[] arr, int i, int j) {
        while(i < j) {
            int tmp = arr[i];
            arr[i] = arr[j];
            arr[j] = tmp;
            i++;
            j--;
        }
    }
}