原文链接: https://leetcode-cn.com/problems/rotate-array

英文原文

Given an array, 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 <= 105
• -231 <= nums[i] <= 231 - 1
• 0 <= k <= 105

 

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?

中文题目

给你一个数组，将数组中的元素向右轮转 k 个位置，其中 k 是非负数。

 

示例 1:

输入: nums = [1,2,3,4,5,6,7], k = 3
输出: [5,6,7,1,2,3,4]
解释:
向右轮转 1 步: [7,1,2,3,4,5,6]
向右轮转 2 步: [6,7,1,2,3,4,5]
向右轮转 3 步: [5,6,7,1,2,3,4]

示例 2:

输入：nums = [-1,-100,3,99], k = 2
输出：[3,99,-1,-100]
解释: 
向右轮转 1 步: [99,-1,-100,3]
向右轮转 2 步: [3,99,-1,-100]

 

提示：

• 1 <= nums.length <= 105
• -231 <= nums[i] <= 231 - 1
• 0 <= k <= 105

 

进阶：

• 尽可能想出更多的解决方案，至少有 三种 不同的方法可以解决这个问题。
• 你可以使用空间复杂度为 O(1) 的 原地 算法解决这个问题吗？

通过代码

高赞题解

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🧠 解题思路

根据题意，如果使用多余数组存储空间，会导致空间复杂度为 $n$，所以在这里，我们可以使用常量级的空间复杂度解法：数组翻转。

思路如下：

1. 首先对整个数组实行翻转，这样子原数组中需要翻转的子数组，就会跑到数组最前面。
2. 这时候，从 $k$ 处分隔数组，左右两数组，各自进行翻转即可。

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🎨 图解演示

<,,,,>

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🍭 示例代码

[]
let reverse = function(nums, start, end){
    while(start < end){
        [nums[start++], nums[end--]] = [nums[end], nums[start]];
    }
}
let rotate = function(nums, k) {
    k %= nums.length;
    reverse(nums, 0, nums.length - 1);
    reverse(nums, 0, k - 1);
    reverse(nums, k, nums.length - 1);
    return nums;
};

[]
class Solution {
public:
    void reverse(vector<int>& nums, int start, int end) {
        while (start < end) {
            swap(nums[start], nums[end]);
            start += 1;
            end -= 1;
        }
    }
    void rotate(vector<int>& nums, int k) {
        k %= nums.size();
        reverse(nums, 0, nums.size() - 1);
        reverse(nums, 0, k - 1);
        reverse(nums, k, nums.size() - 1);
    }
};

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

[]
func reverse(a []int) {
    for i, n := 0, len(a); i < n/2; i++ {
        a[i], a[n-1-i] = a[n-1-i], a[i]
    }
}
func rotate(nums []int, k int) {
    k %= len(nums)
    reverse(nums)
    reverse(nums[:k])
    reverse(nums[k:])
}

[]
void swap(int* a, int* b) {
    int t = *a;
    *a = *b, *b = t;
}
void reverse(int* nums, int start, int end) {
    while (start < end) {
        swap(&nums[start], &nums[end]);
        start += 1;
        end -= 1;
    }
}
void rotate(int* nums, int numsSize, int k) {
    k %= numsSize;
    reverse(nums, 0, numsSize - 1);
    reverse(nums, 0, k - 1);
    reverse(nums, k, numsSize - 1);
}

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转身挥手

嘿，少年，做图不易，留下个赞或评论再走吧！谢啦~ 💐

差点忘了，祝你牛年大吉 🐮 ，AC 和 Offer 📑 多多益善~

⛲⛲⛲ 期待下次再见~

统计信息

通过次数	提交次数	AC比率
383291	857445	44.7%

提交历史

提交时间	提交结果	执行时间	内存消耗	语言

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