Rotate Array

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.

 

第一次提交

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):
        for i in range(0, k):
            nums[i] = nums[i] + len(nums) - k
        for i in range(k, len(nums)):
            nums[i] = nums[i] - k
        print nums

直接给了我红色Error

Runtime Error Message:	Line 7: IndexError: list index out of range
Last executed input:	[-1], 2

index越界，一看就是平时写代码肆意不严格导致的，k比length大就出错了

 

第二次提交，将大于length的部分取模去掉，顺便把array的长度赋给一个值先

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):
        length = len(nums)
        if k > length:
            k = k % length
        for i in range(0, k):
            nums[i] = nums[i] + length - k
        for i in range(k, length):
            nums[i] = nums[i] - k
        print nums

结果貌似没问题，但是还是不完美

Status: Output Limit Exceeded

这里问题意思大概是程序输出结果需要的时间太久，说到底还是程序比较弱，性能不好用时太长，还真想搞两个线程一边一个for，o(∩_∩)o

想着想着，思路也太死板了，实际上也就是元素移位，而我变成了都需要重新计算~！

 

第三次提交，直接用切片合成

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):
        length = len(nums)
        k = k % length
        array1 = nums[:(length - k)]
        array2 = nums[(length - k):]
        nums = array2 + array1
        print nums

本地环境结果正确，可是返回给我的结果说

Status: Wrong Answer

Input:	[1,2], 1
Output:	[1,2]
Expected:	[2,1]

这就奇怪了~！