LeetCode 189

Rotate Array

문제 출처

문제

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
It's guaranteed that nums[i] fits in a 32 bit-signed integer.
k >= 0

나의 코드

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {void} Do not return anything, modify nums in-place instead.
 */
var rotate = function (nums, k) {
  const move = k % nums.length;

  if (nums.length > 1) {
    for (let i = 0; i < nums.length; i++) {
      nums[i] = {
        prev: nums[i],
      };
    }

    for (let i = 0; i < nums.length; i++) {
      let newIndex =
        i - move >= 0 ? i - move : nums.length + ((i - move) % nums.length);
      nums[i] = {
        ...nums[i],
        cur: nums[newIndex].prev,
      };
    }

    for (let i = 0; i < nums.length; i++) {
      nums[i] = nums[i].cur;
    }
  }
};

참고 코드

var rotate = function (nums, k) {
  k %= nums.length; // if k is greater than nums.length then one cycle is completed that means it will remain the same and we have to remainder shifts

  let reverse = function (i, j) {
    while (i < j) {
      let temp = nums[i];
      nums[i] = nums[j];
      nums[j] = temp;
      i++;
      j--;
    }
  }; // suppose  ----->--->
  reverse(0, nums.length - 1); // reverse   <--<------
  reverse(0, k - 1); // reverse first part ---><----
  reverse(k, nums.length - 1); // reverse second part --->----->
};

배운점