189. Rotate Array

Rotate an array of n elements to the right by k steps.

For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].

Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.

Hint:
Could you do it in-place with O(1) extra space?

首先這道題到底是什麼意思？
從字面看，是通過K步來旋轉N個元素的數組。。。
但其實呢，是將一個數組跳過K個數到右邊，也就是將K個數移到開頭

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• 思路1
  在LEETCODE上不能通過 - -！

class Solution {
public:
    vector<int> rotate(vector<int>& nums ,int k) {

    if(k>nums.size())
        k %= nums.size();
    vector<int> v;
    for(int i=nums.size()-k; i<nums.size(); i++)
        v.push_back(nums[i]);
    for(int i=0; i<nums.size()-k; i++)
        v.push_back(nums[i]);

    return v;
}
};

---

• 思路2

很清晰

class Solution {
public:
    vector<int> rotate(vector<int>& nums ,int k) {

    if(k>nums.size())
        k %= nums.size();

    reverse(nums.begin(),nums.end());
    reverse(nums.begin(),nums.begin()+k);
    reverse(nums.begin()+k,nums.end());

    return nums;
}
};