189. 旋转数组 | Rotate 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?

给定一个数组，将数组中的元素向右移动 k 个位置，其中 k 是非负数。

示例 1:

输入: [1,2,3,4,5,6,7] 和 k = 3
输出: [5,6,7,1,2,3,4]
解释:
向右旋转 1 步: [7,1,2,3,4,5,6]
向右旋转 2 步: [6,7,1,2,3,4,5]
向右旋转 3 步: [5,6,7,1,2,3,4]

示例 2:

输入: [-1,-100,3,99] 和 k = 2
输出: [3,99,-1,-100]
解释: 
向右旋转 1 步: [99,-1,-100,3]
向右旋转 2 步: [3,99,-1,-100]

说明:

尽可能想出更多的解决方案，至少有三种不同的方法可以解决这个问题。
要求使用空间复杂度为 O(1) 的原地算法。

 1 class Solution {
 2     func rotate(_ nums: inout [Int], _ k: Int) {
 3         let numsCount = nums.count
 4         if numsCount == 0 || k <= 0
 5         {
 6             return
 7         }
 8         var numsCopy = nums
 9         for i in 0..<numsCount
10         {
11             numsCopy[i] = nums[i]
12         }
13          for j in 0..<numsCount
14         {
15             nums[(j + k)%numsCount] = numsCopy[j]
16         }    
17     }
18 }

16ms

1 class Solution {
2     func rotate(_ nums: inout [Int], _ k: Int) {
3         let index = nums.count - 1 - (k % nums.count)
4         nums += nums[0...index]
5         nums.removeSubrange((0...index))
6     }
7 }

20ms

1 class Solution {
2     func rotate(_ nums: inout [Int], _ k: Int) {
3         let realK = k % nums.count
4         let sourceNums = nums
5         for i in 0 ..< sourceNums.count {
6             nums[(i+k) % sourceNums.count] = sourceNums[i]
7         }
8     }
9 }

12ms

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 1 class Solution {
 2     func rotate(_ nums: inout [Int], _ k: Int) {
 3         guard nums.count > 1 else {
 4             return
 5         }
 6         
 7         guard k > 0 else {
 8             return
 9         }
10         
11         let k = k % nums.count        
12         nums = Array(nums[(nums.count - k)..<nums.count]) + Array(nums[0..<(nums.count - k)])
13         
14         // O(k % nums.count) extra space
15         /*
16         let shiftOffset = k % nums.count
17         let temp = Array(nums[(nums.count - shiftOffset)..<nums.count])
18         for i in (shiftOffset..<nums.count).reversed() {
19             nums[i] = nums[i - shiftOffset]
20         }
21         
22         for i in 0..<shiftOffset {
23             nums[i] = temp[i]
24         }*/
25         
26     }
27 }

189. 旋转数组 | Rotate Array

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