leetcode 189. Rotate Array

/*leetcode 189. Rotate Array题目大意：Rotate an array of n elements to the right by k steps.譬如说对于k=3,[1,2,3,4,5,6,7]旋转为[5,6,7,1,2,3,4].解题思路：用一个临时的数组，然后根据k的值,旋转临时数组，然后复制到原来的数组*/#include <iostream>#include <vector>#include <string>using namespace std;class Solution {public:    //方法1：使用临时数组，先复制后k个元素，再负责n-7个元素    void rotate(vector<int>& nums, int k)     {        int n = nums.size();        k = k % n;        if (n <= 1 || k == 0)            return;        vector<int> tmp;        for (int i = n -k ; i < n; ++i)            tmp.push_back(nums[i]);        for (int i = 0;i < n-k; ++i)            tmp.push_back(nums[i]);        nums = tmp;    }    //方法2：直接使用vector的函数    void rotate2(vector<int>& nums, int k)    {        int n = nums.size();        k = k % n;        if (n <= 1 || k == 0)            return;        vector<int> extra(nums.begin(), nums.begin() + n - k);        //for (auto i : extra)        //  cout << i << " ";        //cout << endl;        nums.erase(nums.begin(), nums.begin()+n-k);        //for (auto i : nums)        //  cout << i << " ";        //cout << endl;        nums.insert(nums.end(), extra.begin(), extra.end());    }};void test_rotate(){    vector<int> nums{ 1, 2 ,3, 4, 5, 6, 7};    Solution sol;    sol.rotate(nums, 3);    for (auto i : nums)        cout << i << " ";    cout << endl;}void test_rotate2(){    vector<int> nums{ 1, 2 ,3, 4, 5, 6, 7 };    Solution sol;    sol.rotate2(nums, 3);    for (auto i : nums)        cout << i << " ";    cout << endl;}int main(){    //test_rotate();    test_rotate2();    return 0;}