leetcode189:旋转数组

思想:

首先判断nums是否为空以及k是否为零,若是则直接返回nums。然后定义一个反转函数reverse,使得数组前后元素调换位置,直到全部调换。最后先nums[0,len(nums)-1-k]开始调换,然后nums[en(nums)-1-k+1,len(nums)-1]开始调换,最后nums[0,len(nums)-1]调换。

class Solution:
    def rotate(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: void Do not return anything, modify nums in-place instead.
        """
        if len(nums) == 0 or k == 0:
            return
        def reverse(start, end, s):
            while start < end:
                s[start], s[end] = s[end], s[start]
                start += 1
                end -= 1
                
        n = len(nums) - 1
        k = k % len(nums)
        reverse(0, n - k, nums)
        reverse(n - k + 1, n, nums)
        reverse(0, n, nums)

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转载自blog.csdn.net/weixin_43160613/article/details/83576690