Leetcode 189. 轮转数组

原题链接：Leetcode 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 <= 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?

题目大意：

给定一个整数数组 nums，将数组中的元素向右轮转 k 个位置。
题目有多种解法。

方法一：使用额外数组

思路：

最先想到的笨办法，将元素移动到临时数组对应的位置，最后再将元素取回原数组

C++ 代码：

class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> tmp(n);

        for(int i = 0; i < n; i++ ){
            tmp[(i + k) % n] = nums[i];
        }

        nums.assign(tmp.begin(), tmp.end());
    }
};

复杂度分析：

• 时间复杂度：O(n)，遍历了一遍数组
• 空间复杂度：O(n)，用了一个和原数组等长的临时数组

补充：vector容器的assign():

函数原型：

• void assign(const_iterator first,const_iterator last);
• void assign(size_type n,const T& x = T());

功能：

• 将区间 [first,last) 的元素赋值到当前的vector容器中，或者n个值为x的元素到vector容器中，这个容器会清除掉vector容器中以前的内容。