[LeetCode] 189. Rotate Array_Easy tag: Array

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?

这个题目要用“三步翻转法”

对于一个[1, 2, 3, 4, 5, 6, 7] 这样的数组来说的话

#rotate 3 steps
# . 1 2 3 4    5 6 7 , k = 3
# . 4 3 2 1    7 6 5
#   5 6 7     1 2 3 4   rotate whole array

另外有个重要的点是自己写一个in place的reverse的helper 函数

def reverse(nums, start, end):
    while start < end:
        nums[start], nums[end] = nums[end], nums[start]
        start += 1
        end -= 1

 

Code

class Solution:
    def rotateArray(self, nums, k):
        """
        Do not return anything, modify nums in-place instead.
        """
        if not k or not nums: return
        n = len(nums)
        k = k%n   # we just need to do k%n times, save time
        def reverse(nums, start, end):
            while start < end:
                nums[start], nums[end] = nums[end], nums[start]
                start += 1
                end -= 1
        reverse(nums, 0, n - k - 1) # reverse the first part
        reverse(nums, n - k, n - 1) # reverse the last part
        reverse(nums, 0, n - 1) # reverse the whole array