20170607-leetcode-189-Rotate Array

1.Description

Rotate an array of n elements to the right by k steps.

For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].

Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.

[show hint]

Related problem: Reverse Words in a String IIRotate an array of n elements to the right by k steps.

For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].

Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.

[show hint]

Related problem: Reverse Words in a String II
解读
把list（或者array）看成是一个环，求向右转动k个数之后的list
比如1,2,3,4,5,6,7，想象7指向1：

向右移动1，结果为：7,1,2,3,4,5,6向右移动2，结果为：6,7,1,2,3,4,5向右移动3，结果为：5,6,7,1,2,3,4

2.Solution

2.1找到分界点,断开，重新组合，给list重新赋值

class Solution(object):    def rotate(self, nums, k):        n = len(nums)        k = k % n        nums[:] = nums[n-k:] + nums[:n-k]#或者简写成下面的形式：def rotate(self, nums, k):        k = k % len(nums)        nums[:] = nums[-k:] + nums[:-k]

注：最后的写成是num[:]而不是num的原因是，用num进行赋值会改变原来的值，而num[:]只是修改指针，效率更高

2.2

以n - k为界，分别对数组的左右两边执行一次逆置；然后对整个数组执行逆置。

class Solution:    # @param nums, a list of integer    # @param k, num of steps    # @return nothing, please modify the nums list in-place.    def rotate(self, nums, k):        n = len(nums)        k %= n        self.reverse(nums, 0, n - k)        self.reverse(nums, n - k, n)        self.reverse(nums, 0, n)    def reverse(self, nums, start, end):        for x in range(start, (start + end) / 2):            nums[x] ^= nums[start + end - x - 1]#交换赋值            nums[start + end - x - 1] ^= nums[x]#交换赋值            nums[x] ^= nums[start + end - x - 1]

注：

Python中两个数交换可以用如下方法实现：a, b = b, a或者：a ^= bb ^= aa ^= b

2.3不断移动元素进行实现

将数组元素依次循环向右平移k个单位

class Solution:    def rotate(self, nums, k):        n = len(nums)        idx = 0        distance = 0        cur = nums[0]        for x in range(n):            idx = (idx + k) % n            nums[idx], cur = cur, nums[idx]            distance = (distance + k) % n            if distance == 0:                idx = (idx + 1) % n                cur = nums[idx]

2.4数组(list)长度加倍，截断

class Solution(object):    def rotate(self, nums, k):        n=len(nums)        k = k % n        nums[:]=(nums*2)[n-k:2*n-k]

2.5下面一种方式超时

    def rotate(self, nums, k):        k = k % len(nums)        for i in range(k):            nums[:] = [nums[-1]] + nums[:-1]