189. Rotate Array

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
  還沒看討論區的答案，思路就是先將陣列中最後一個數字儲存在temp中，再通過迴圈將索引為0到size()-1的元素向右移，最後將temp的值賦給v[0].

class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        if(k == 0)
            return;
        for(int i = 0; i<k; i++)
        {
            int temp = nums[nums.size()-1];
            for(int i = nums.size() -1; i>0; i--)
            {
                nums[i] = nums[i-1];
            }
            nums[0] = temp;
        }
    }
};