相似的Leetcode題目：
Sort Array By Parity

Rotate Array題目：
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?
 

解題思路：題目是對一陣列中的最後一個元素從尾部向頭部進行移動，後面k個元素往前移動和前面n-k個元素進行拼接。

def Solution(array, k):
    if len(array) == 0 or array is None:
        return array
    if k<=len(array):
        return array[-k:] + array[:-k]
    if k>len(array):
        k = k % len(array)
        return array[-k:] + array[:-k]
 

Solution中給到了最優解法（時間複雜度O(n)，空間複雜度O(1)）, 具體解法見：
解法
第三種解法沒看懂，第四種解法可以參考：

Original List                   : 1 2 3 4 5 6 7
After reversing all numbers     : 7 6 5 4 3 2 1
After reversing first k numbers : 5 6 7 4 3 2 1
After revering last n-k numbers : 5 6 7 1 2 3 4 --> Result

java

public class Solution {
    public void rotate(int[] nums, int k) {
        k %= nums.length;
        reverse(nums, 0, nums.length - 1);
        reverse(nums, 0, k - 1);
        reverse(nums, k, nums.length - 1);
    }
    public void reverse(int[] nums, int start, int end) {
        while (start < end) {
            int temp = nums[start];
            nums[start] = nums[end];
            nums[end] = temp;
            start++;
            end--;
        }
    }
}

Complexity Analysis

    Time complexity : O(n). nnn elements are reversed a total of three times.

    Space complexity : O(1). No extra space is used.

Ref:
1、https://leetcode.com/problems/rotate-array/description/