【LeetCode】【Array】rotate array

Question

给定一个数组(array)，将数组向右旋转k步(k>=0)。

不需要输出，只改变原来的数组就可以。

Solution 1:

每次循环将数组的元素都向后移动一位，循环k次。

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        for i in range(k):
            previous = nums[-1]
            for j in range(len(nums)):
                nums[j], previous = previous, nums[j]

nums[j], previous = previous, nums[j]中，nums[j]和previous都是原先的值；

nums[j] = previous

previous = nums[j]

中，nums[j]的值改变成previous的值了，所以两者的结果不同。

• Time complexity: O(n×k)\mathcal{O}(n \times k)O(n×k). All the numbers are shifted by one step(O(n)\mathcal{O}(n)O(n)) k times(O(n)\mathcal{O}(n)O(n)).

• Space complexity: O(1)\mathcal{O}(1)O(1). No extra space is used.

Solution 2:

构建一个新的数组，并将元素直接移动到正确的位置。

用求余的方式来实现错位。

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        n = len(nums)
        a = [0] * n
        for i in range(n):
            a[(i + k) % n] = nums[i]  #%求余
            
        nums[:] = a

• Time complexity: O(n)\mathcal{O}(n)O(n). One pass is used to put the numbers in the new array. And another pass to copy the new array to the original one.

• Space complexity: O(n)\mathcal{O}(n)O(n). Another array of the same size is used.

Solution 3

循环替换

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        n = len(nums)
        k %= n
        
        start = count = 0
        while count < n:
            current, prev = start, nums[start]
            while True:
                next_idx = (current + k) % n
                nums[next_idx], prev = prev, nums[next_idx]
                current = next_idx
                count += 1
                
                if start == current:
                    break
            start += 1

• Time complexity: O(n)\mathcal{O}(n)O(n). Only one pass is used.

• Space complexity: O(1)\mathcal{O}(1)O(1). Constant extra space is used.

Solution 4:

使用列表reverse

class Solution:
    def reverse(self, nums: list, start: int, end: int) -> None:
        while start < end:
            nums[start], nums[end] = nums[end], nums[start]
            start, end = start + 1, end - 1
                
    def rotate(self, nums: List[int], k: int) -> None:
        n = len(nums)
        k %= n

        self.reverse(nums, 0, n - 1)
        self.reverse(nums, 0, k - 1)
        self.reverse(nums, k, n - 1)

• Time complexity: O(n)\mathcal{O}(n)O(n). nnn elements are reversed a total of three times.

• Space complexity: O(1)\mathcal{O}(1)O(1). No extra space is used.