189.rotate-array.cpp

//  Tag: Array, Math, Two Pointers
//  Time: O(N)
//  Space: O(1)
//  Ref: -
//  Note: -

//  Given an integer array nums, rotate the array to the right by k steps, where k is non-negative.
//   
//  Example 1:
//  
//  Input: nums = [1,2,3,4,5,6,7], 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: nums = [-1,-100,3,99], 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]
//  
//   
//  Constraints:
//  
//  1 <= nums.length <= 105
//  -231 <= nums[i] <= 231 - 1
//  0 <= k <= 105
//  
//   
//  Follow up:
//  
//  Try to come up with as many solutions as you can. There are at least three different ways to solve this problem.
//  Could you do it in-place with O(1) extra space?
//  
//  

class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> rotated(n);
        for (int i = 0; i < n; ++i) {
            rotated[(i + k) % n] = nums[i];
        }
        nums = rotated;
    }   
};

class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        int n = nums.size();
        k = k % n;
        if (k == 0) {
            return;
        }
        reverse(nums, 0, n - 1);
        reverse(nums, 0, k - 1);
        reverse(nums, k, n - 1);
    }   

    void reverse(vector<int>& nums, int left, int right) {
        while (left < right) {
            swap(nums[left++], nums[right--]);
        }

    }
};