[Swift]LeetCode189. 旋转数组 | Rotate Array

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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。

---

48ms

class Solution {
    func rotate(_ nums: inout [Int], _ k: Int) {
        guard nums.count > 1 else {
            return
        }
        
        guard k > 0 else {
            return
        }
        
        let k = k % nums.count        
        nums = Array(nums[(nums.count - k)..<nums.count]) + Array(nums[0..<(nums.count - k)])        
    }
}

---

48ms

class Solution {
    func rotate(_ nums: inout [Int], _ k: Int) {
         let n = nums.count
        let numRoations = k % n
        var rotated = [Int]()
        for i in n-numRoations..<n {
            rotated.append(nums[i])
        }
        for j in 0..<n-numRoations {
            rotated.append(nums[j])
        }
        nums = rotated
    }
}

---

52ms

class Solution {
    func rotate(_ nums: inout [Int], _ k: Int) {
        rotate1(&nums, k)
    }
    
    func rotate1(_ nums: inout [Int], _ k: Int) {
        var r: [Int] = Array(repeating: 0, count: nums.count)
        var kk = nums.count - (k % nums.count)
        for i in 0..<nums.count {
            r[i] = nums[(i + kk) % nums.count]
        }
        for (i, num) in r.enumerated() {
            nums[i] = num
        }
    }
}

---

56ms

class Solution {
    func rotate(_ nums: inout [Int], _ k: Int) {
        guard nums.count > 0, k > 0 else {
            return
        }
        
        let k = k%nums.count
        
        rotate(&nums, 0, nums.count-k-1)
        rotate(&nums, nums.count-k, nums.count-1)
        rotate(&nums, 0, nums.count-1)
    }
    
    private func rotate(_ nums: inout [Int], _ left: Int, _ right: Int) {
        if left == right || nums.count == 0 { return }
        
        var i = left, j = right
        
        while i < j {
            let temp = nums[i]
            nums[i] = nums[j]
            nums[j] = temp
            i += 1
            j -= 1
        }
    }
}

---

64ms

class Solution {
    func rotate(_ nums: inout [Int], _ k: Int) {
        var n = nums.count
        let ind = n-1-k%n
        nums += nums[0...ind]
        nums.removeSubrange(0...ind)
    }
}