Rotate Array - LeetCode

https://leetcode.com/problems/rotate-array/?envType=study-plan-v2&envId=top-interview-150

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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?

解析

简单的方法是直接pop然后放在首位，使用insert。代码很短，显然不是最优解。

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        while k > 0:
            nums.insert(0, nums.pop())
            k -= 1

另一种解法是翻转数组达到效果。

For nums = [1,2,3,4,5,6,7] and k = 3:

  1. Reverse the whole array: [7,6,5,4,3,2,1]
  2. Reverse the first k elements: [5,6,7,4,3,2,1]
  3. Reverse the remaining elements: [5,6,7,1,2,3,4]

因此需要写一个reverse方法。

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        n = len(nums)
        k %= n
        
        def reverse(l, start, end):
            while start < end:
                l[start], l[end] = l[end], l[start]
                start += 1
                end -= 1
        reverse(nums, 0, n - 1)
        reverse(nums, 0, k - 1)
        reverse(nums, k, n - 1)

上面的方法runtime少了10倍。

class Solution {
    public void rotate(int[] nums, int k) {
       int n = nums.length;
        k %= n;

        reverse(nums, 0, n - 1);
        reverse(nums, 0, k - 1);
        reverse(nums, k, n - 1);

    }

    public void reverse(int[] l, int start, int end){
        while(start < end){
            int temp;
            temp = l[start];
            l[start] = l[end];
            l[end] = temp;
            start++;
            end--;

        }
    }
}

class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        int n = nums.size();
        k %= n;

        reverse(nums, 0 ,n - 1);
        reverse(nums, 0, k - 1);
        reverse(nums, k, n - 1);

    }
    void reverse(vector<int>& nums, int start, int end){
        while (start < end){
            int temp;
            temp = nums[start];
            nums[start] = nums[end];
            nums[end] = temp;
            start++;
            end--;
        }
    }
};