Leetcode 189. Rotate Array vector取最后一个元素，删除erase()用法

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

题目链接:https://leetcode.com/problems/rotate-array/

class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        while(k>0)
        {
            int t=nums.back();
            nums.pop_back();
            nums.insert(nums.begin(),t);
            k--;
        }
    }
};

复习一下vector的erase()用法

#include<bits/stdc++.h>
using namespace std;

int main()
{
    vector<int> v;
    for(int i=0;i<6;i++)
        v.emplace_back(i);
    v.erase(v.begin(),v.begin()+2);
    for(int i=0;i<v.size();i++)
        cout<<v[i]<<endl;
   return 0;
}