[每日 LeetCode] 189. Rotate Array

Description:

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?

思路：题目要求将数组旋转 k 次。可以先创建一个新数组，然后将原数组中的后 k 个元素放到新数组的前 k 个位置，将原数组中的前 n-k 个元素放到新数组的后 n-k 个位置，然后将新数组的元素复制到旧数组中，最后删除新数组。

注意：c++ 中 memcpy 函数用法

函数原型： void *memcpy(void*dest, const void *src, size_t n);

功能：由src指向地址为起始地址的连续n个字节的数据复制到以destin指向地址为起始地址的空间内。

C++ 代码

class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        int n=nums.size();
        k=k%n;
        int* temp=new int[n];
        memcpy(temp,&nums[0]+(n-k),sizeof(int)*k);
        memcpy(temp+k,&nums[0],sizeof(int)*(n-k));
        memcpy(&nums[0],temp,sizeof(int)*n);
        delete[] temp;
    }
};

运行时间：16ms
运行内存：