Description
Given an array, rotate the array to the right by k steps, where k is non-negative.
Example 1:
Input: [1,2,3,4,5,6,7] and 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: [-1,-100,3,99] and 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]
描述
给定一个数组，将数组中的元素向右移动 k 个位置，其中 k 是非负数。
示例 1:
输入: [1,2,3,4,5,6,7] 和 k = 3
输出: [5,6,7,1,2,3,4]
解释:
向右旋转 1 步: [7,1,2,3,4,5,6]
向右旋转 2 步: [6,7,1,2,3,4,5]
向右旋转 3 步: [5,6,7,1,2,3,4]
示例 2:
输入: [-1,-100,3,99] 和 k = 2
输出: [3,99,-1,-100]
解释:
向右旋转 1 步: [99,-1,-100,3]
向右旋转 2 步: [3,99,-1,-100]
思路
- 先将整理数组倒转一次.
- 再将数组前k个元素倒转一次.
- 最后将k+1到嘴后一个元素倒转一次.
# -*- coding: utf-8 -*- # @Author: 何睿 # @Create Date: 2019-01-19 17:54:06 # @Last Modified by: 何睿 # @Last Modified time: 2019-01-19 18:26:39 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. """ k %= len(nums) self.reverse(nums, 0, len(nums) - 1) self.reverse(nums, 0, k - 1) self.reverse(nums, k, len(nums) - 1) def reverse(self, nums, start, end): while start < end: nums[start], nums[end] = nums[end], nums[start] start, end = start + 1, end - 1
源代码文件在这里.
