[LeetCode] 189. Rotate Array

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?

轮转数组。

给定一个整数数组 nums，将数组中的元素向右轮转 k 个位置，其中 k 是非负数。

思路

题意跟61题几乎一样，唯一不同的是这题的 input 是数组，所以代码上没什么参考性。这个题目的思路很巧妙，叫做三步反转法。我就拿第一个例子遍历好了。数组长度为7，要往右 rotate 三步。三步反转法的思路是

start, [1, 2, 3, 4, 5, 6, 7]
整个rotate，[7, 6, 5, 4, 3, 2, 1]
rotate前k个， [5, 6, 7, 4, 3, 2, 1]
rotate后n - k个，[5, 6, 7, 1, 2, 3, 4]

复杂度

时间O(n)
空间O(1)

代码

Java实现

class Solution {
    public void rotate(int[] nums, int k) {
        k = 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--;
        }
    }
}

JavaScript实现

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {void} Do not return anything, modify nums in-place instead.
 */
var rotate = function(nums, k) {
    k = k % nums.length;
    reverse(nums, 0, nums.length - 1);
    reverse(nums, 0, k - 1);
    reverse(nums, k, nums.length - 1);
};

var reverse = function(nums, start, end) {
    while (start < end) {
        let temp = nums[start];
        nums[start] = nums[end];
        nums[end] = temp;
        start++;
        end--;
    }
};

还有一种JS的思路是 pop 再 shift，利用的是 JS 对 array 的各项操作。
时间O(k)
空间O(n)

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {void} Do not return anything, modify nums in-place instead.
 */
var rotate = function(nums, k) {
    k = k % nums.length;
    for (var i = 1; i <= k; i++) {
        nums.unshift(nums.pop());
    }
};

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