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rotate-array.py
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rotate-array.py
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'''
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]
Note:
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?
'''
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.
"""
# Approach one 最简单的思路,在数组后面先补出来k个数字,然后数组向后移动,再把后面的数字搬到前面,最后去掉后面多余的数字。效率低下的算法
# length = len(nums)
# for i in range(k):
# nums.append(0)
# for i in range(len(nums)-k-1,-1,-1):
# nums[i+k] = nums[i]
# for j in range(len(nums)-1,len(nums)-k-1,-1):
# nums[k-1] = nums[j]
# k -= 1
# for j in range(len(nums)-length):
# del nums[-1]
# Approach Two
# while k:
# k -= 1
# nums.insert(0, nums.pop(-1))
# Approach Three:运用数组的切片操作,简洁。并且是一个原地算法。
i = k % len(nums)
nums[:] = nums[-i:]+nums[:-i]