189.rotate-array.cpp

/*
 * @lc app=leetcode id=189 lang=cpp
 *
 * [189] Rotate Array
 *
 * https://leetcode.com/problems/rotate-array/description/
 *
 * algorithms
 * Medium (36.32%)
 * Likes:    4224
 * Dislikes: 911
 * Total Accepted:    652K
 * Total Submissions: 1.8M
 * Testcase Example:  '[1,2,3,4,5,6,7]\n3'
 *
 * Given an array, rotate the array to the right by k steps, where k is
 * non-negative.
 * 
 * Follow up:
 * 
 * 
 * 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?
 * 
 * 
 * 
 * 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 <= 2 * 10^4
 * -2^31 <= nums[i] <= 2^31 - 1
 * 0 <= k <= 10^5
 * 
 * 
 */

// @lc code=start
class Solution {
public:
    void rotate1(vector<int>& nums, int k) {
        k = k % nums.size();
        int left = nums.size() - k - 1;
        int right = nums.size() - 1;
        reverse(nums, 0, left);
        reverse(nums, left + 1, right);
        reverse(nums, 0, right);
    }
    
    void reverse(vector<int>& nums, int left, int right) {
        while (left < right) {
            swap(nums[left++], nums[right--]);
        }
    }
    void rotate(vector<int>& nums, int k) {
        k = k % nums.size();
        int count = 0;
        int n = nums.size();
        for (int start = 0; count < nums.size(); ++start) {
            int current = start;
            int prev = nums[start];
            do {
                int next = (current + k) % n;
                swap(prev, nums[next]);
                current = next;
                ++count;
            } while (start != current);
        }
    }
};
// @lc code=end