LeetCode 189. 轮转数组

189. Rotate Array

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

Example 1:

Input: nums = [1,2,3,4,5,6,7], 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: nums = [-1,-100,3,99], 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]

Constraints:

• 1 <= nums.length <= 10^5
• -2^31 <= nums[i] <= 2^31 - 1
• 0 <= k <= 10^5

Follow up:

Try to come up with as many solutions as you can. There are at least three different ways to solve this problem.

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

解法思路：

1、暴力求解

2、双指针（法一）

3、递归翻转（法二）

?法一：

class Solution {
    public void rotate(int[] nums, int k) {
        // double pointer
        // Time: O(n)
        // Space: O(n)
        int n = nums.length;
        int[] newArr = new int[n];
        for (int i = 0; i < n; i++) {
            newArr[(i + k) % n] = nums[i];
        }
        System.arraycopy(newArr, 0, nums, 0, n);
    }
}

?法二：

class Solution {
    public void rotate(int[] nums, int k) {
        // Recursion
        // Time: O(n)
        // Space: O(1)
        k %= nums.length;
        reverse(nums, 0, nums.length - 1);
        reverse(nums, 0, k - 1);
        reverse(nums, k, nums.length - 1);
    }

    private void reverse(int[] nums, int start, int end) {
        while (start < end) {
            int temp = nums[start];
            nums[start++] = nums[end];
            nums[end--] = temp;
            // ++start;
            // --end;
        }
    }
}

?