LeetCode/189.rotate-array.cpp at master · samba9274/LeetCode · GitHub

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/*
 * @lc app=leetcode id=189 lang=cpp
 *
 * [189] Rotate Array
 *
 * https://leetcode.com/problems/rotate-array/description/
 *
 * algorithms
 * Medium (38.87%)
 * Likes:    10817
 * Dislikes: 1386
 * Total Accepted:    1.2M
 * Total Submissions: 3.2M
 * 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.
 *
 *
 * 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 <= 10^5
 * -2^31 <= nums[i] <= 2^31 - 1
 * 0 <= k <= 10^5
 *
 *
 *
 * 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?
 *
 *
 */

// @lc code=start
class Solution
{
public:
    void rotate(vector<int> &nums, int k)
    {

        k %= nums.size();
        int n = nums.size();
        for (int i = n - k; i < n; i++)
            nums.push_back(nums[i]);
        for (int i = n - k - 1, j = n - 1; i >= 0; i--, j--)
            nums[j] = nums[i];
        for (int i = n, j = 0; i < n + k; i++, j++)
            nums[j] = nums[i];
        nums.erase(nums.begin() + n, nums.end());
    }
};
// @lc code=end