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?
解题要点：
记录第一段和第二段，放在一个temp数组里，然后遍历把temp的数放到nums。
class Solution(object):
def rotate(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: None Do not return anything, modify nums in-place instead.
"""
k = k % len(nums)
firstHalf = []
for i in range(len(nums)-k):
firstHalf.append(nums[i])
secondHalf = []
for j in range(len(nums)-k, len(nums)):
secondHalf.append(nums[j])
temp = secondHalf + firstHalf
for i in range(len(nums)):
nums[i] = temp[i]
In-place && O(1) extra memory 解法：
class Solution(object):
def rotate(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: None Do not return anything, modify nums in-place instead.
"""
def swap(nums, i, j):
while i < j:
temp = nums[i]
nums[i] = nums[j]
nums[j] = temp
i += 1
j -=1
k = k % len(nums)
swap(nums, 0, len(nums)-1)
swap(nums, 0, k - 1)
swap(nums, k, len(nums)-1)