[Swift]LeetCode31. 下一个排列 | Next Permutation

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Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.

If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).

The replacement must be in-place and use only constant extra memory.

Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.

1,2,3 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1

---

实现获取下一个排列的函数，算法需要将给定数字序列重新排列成字典序中下一个更大的排列。

如果不存在下一个更大的排列，则将数字重新排列成最小的排列（即升序排列）。

必须原地修改，只允许使用额外常数空间。

以下是一些例子，输入位于左侧列，其相应输出位于右侧列。
1,2,3 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1

---

 【一遍扫描】算法

首先，我们观察到对于任何给定序列的降序，没有可能的下一个更大的排列。

例如，以下数组不可能有下一个排列：

[9, 5, 4, 3, 1]

我们需要从右边找到第一对两个连续的数字 a[i]a[i] 和 a[i-1]a[i−1]，它们满足 a[i]>a[i-1]a[i]>a[i−1]。现在，没有对 a[i-1]a[i−1]右侧的重新排列可以创建更大的排列，因为该子数组由数字按降序组成。因此，我们需要重新排列 a[i-1]a[i−1] 右边的数字，包括它自己。

现在，什么样的重新排列将产生下一个更大的数字？我们想要创建比当前更大的排列。因此，我们需要将数字 a[i-1]a[i−1] 替换为位于其右侧区域的数字中比它更大的数字，例如 a[j]a[j]。

我们交换数字 a[i-1]a[i−1] 和 a[j]a[j]。我们现在在索引 i-1i−1 处有正确的数字。 但目前的排列仍然不是我们正在寻找的排列。我们需要通过仅使用 a[i-1]a[i−1]右边的数字来形成最小的排列。 因此，我们需要放置那些按升序排列的数字，以获得最小的排列。

但是，请记住，在从右侧扫描数字时，我们只是继续递减索引直到我们找到 a[i]a[i] 和 a[i-1]a[i−1] 这对数。其中，a[i] > a[i-1]a[i]>a[i−1]。因此，a[i-1]a[i−1] 右边的所有数字都已按降序排序。此外，交换 a[i-1]a[i−1] 和 a[j]a[j] 并未改变该顺序。因此，我们只需要反转 a[i-1]a[i−1] 之后的数字，以获得下一个最小的字典排列。

下面的动画将有助于你理解：

---

---

 24ms

 1 class Solution {
    func nextPermutation(_ nums: inout [Int]) {
        let len:Int = nums.count
        var i:Int = len - 2
        while (i >= 0 && nums[i + 1] <= nums[i]) {i -= 1}
        if i >= 0
        {
            var j:Int = len - 1
            while (j >= 0 && nums[j] <= nums[i]) {j -= 1}
             swap(&nums, i, j)
        }
        reverse(&nums, i + 1)
    }
    
    func reverse(_ nums:inout [Int],_ start:Int)
    {
        var i:Int = start, j:Int = nums.count - 1
        while (i < j)
        {
            swap(&nums, i, j)
            i += 1
            j -= 1
        }
    }
    
    func swap(_ nums:inout [Int],_ i:Int,_ j:Int)
    {
        var temp:Int = nums[i]
        nums[i] = nums[j]
        nums[j] = temp
    }
}

---

12ms

class Solution {
    func nextPermutation(_ nums: inout [Int]) {
        if nums.count <= 1 { return }
        var ind = -1
        for i in (0..<(nums.count-1)).reversed() {
            if nums[i] < nums[i+1] {
                ind = i
                break
            }
        }
        if ind == -1 { 
            nums = nums.reversed() 
            return
        }
        let ind1 = ind
        var temp = nums[ind+1]
        var ind2 = ind+1
        while ind < nums.count {
            if nums[ind] > nums[ind1] && nums[ind] <= temp { 
                ind2 = ind
                temp = nums[ind]
            }
            ind += 1
        }
        (nums[ind1], nums[ind2]) = (nums[ind2], nums[ind1])
        nums = Array(nums[0...ind1]) + Array(nums[(ind1+1)..<nums.count]).reversed()
    }
}

---

16ms

class Solution {
    func nextPermutation(_ nums: inout [Int]) {
        guard let violateIndex = findViolate(nums) else {
            nums.reverse()
            return
        }
        
        swap(&nums, violateIndex, findLeastGreater(nums, violateIndex))
        nums = nums[0...violateIndex] + nums[(violateIndex + 1)...].reversed()
    }
    
    fileprivate func findViolate(_ nums: [Int]) -> Int? {
        for i in (1..<nums.count).reversed() {
            if nums[i] > nums[i - 1] {
                return i - 1
            }
        }
        
        return nil
    }
    
    fileprivate func findLeastGreater(_ nums: [Int], _ violateIndex: Int) -> Int {
        for i in (violateIndex + 1..<nums.count).reversed() {
            if nums[i] > nums[violateIndex] {
                return i
            }
        }
        
        fatalError()
    }
    
    fileprivate func swap<T>(_ nums: inout [T], _ indexL: Int, _ indexR: Int) {
        (nums[indexL], nums[indexR]) = (nums[indexR], nums[indexL])
    }
}

---

16ms

class Solution {
    func nextPermutation(_ nums: inout [Int]) {
        
        guard nums.count > 1 else {
            return
        }
        
        for i in (0 ..< nums.count - 1).reversed() {
            
            var lowestDiff = -1
            var lowestIndex = -1
            
            for j in (i+1 ..< nums.count) {
               
                let diff = nums[j] - nums[i]
                
                if diff > 0 && (diff < lowestDiff || lowestDiff == -1) {
                    lowestDiff = diff
                    lowestIndex = j
                }
            }
            
            if (lowestIndex >= 0) {
                print("swap \(nums[i]) \(nums[lowestIndex])")
                swap(&nums, i, lowestIndex)
                nums[i+1..<nums.count].sort()
                return
            }
        }
        
        for i in (0 ..< nums.count / 2) {
            swap(&nums, i, nums.count - i - 1)
        }
    }
    
    func swap(_ nums: inout [Int], _ i: Int, _ j: Int) {
        let tmp = nums[i]
        nums[i] = nums[j]
        nums[j] = tmp
    }
}

---

20ms

class Solution {
    func nextPermutation(_ nums: inout [Int]) {
        
        var end = nums.count - 1
       
        while end > 0 {
            if nums[end - 1] < nums[end] {
                break
            }
            end -= 1
        }
        
        if end == 0 {
            for i in 0..<nums.count / 2 {
                nums.swapAt(i, nums.count - 1 - i)
            }
        } else {
            end -= 1
            var i = nums.count - 1
            while i > end {
                if nums[i] > nums[end] {
                    break
                }
                i -= 1
            }
            
            nums.swapAt(i, end)
            
            for i in 0..<(nums.count - end) / 2 {
                nums.swapAt(end + 1 + i, nums.count - 1 - i)
            } 
        }
    }
}

---

20ms

class Solution {
    func nextPermutation(_ nums: inout [Int]) {
        
        let size = nums.count
        
        //数组为空或只有一个数
        if size == 0 || size == 1 {
            return
        }
        
        //只有两个数直接交换
        if size == 2 {
            nums.swapAt(0, 1)
            return
        }
        
        //判断大小交换位置
        for index in stride(from: size-1, through: 1, by: -1) {
            
            if nums[index] > nums[index - 1] {
                
                for subIndex in stride(from: size-1, through: index, by: -1) {
                    
                    if nums[subIndex] > nums[index-1] {
                        
                        nums.swapAt(index-1, subIndex)
                        break
                    }
                }
                
                for tmpIndex in stride(from: (size-1-index)/2, through: 0, by: -1) {
                    
                    nums.swapAt(tmpIndex + index, size - 1 - tmpIndex)
                }
                
                return
                
            }
        }
        
        //降序排列好,直接逆序
        for index in stride(from: (size-1)/2, through: 0, by: -1) {
            
            nums.swapAt(index, size-1-index)
        }
    }
}

---

24ms

class Solution {
    func nextPermutation(_ nums: inout [Int]) {
        
        guard nums.count > 1 else {
            return
        }
        
        for i in (0 ..< nums.count - 1).reversed() {
            
            var lowestDiff = -1
            var lowestIndex = -1
            
            for j in (i+1 ..< nums.count) {
               
                let diff = nums[j] - nums[i]
                
                if diff > 0 && (diff < lowestDiff || lowestDiff == -1) {
                    lowestDiff = diff
                    lowestIndex = j
                }
            }
            
            if (lowestIndex >= 0) {
                print("swap \(nums[i]) \(nums[lowestIndex])")
                swap(&nums, i, lowestIndex)
                nums[i+1..<nums.count].sort()
                return
            }
        }
        
        for i in (0 ..< nums.count / 2) {
            swap(&nums, i, nums.count - i - 1)
        }
    }
    
    func swap(_ nums: inout [Int], _ i: Int, _ j: Int) {
        let tmp = nums[i]
        nums[i] = nums[j]
        nums[j] = tmp
    }
}

---

28ms

class Solution {
    func nextPermutation(_ nums: inout [Int]) {
        if nums.count < 2 {
            return
        }
        
        var max = nums[nums.count - 1]
        var i = nums.count - 2
        while i >= 0 {
            if nums[i] < max {
                break
            }
            max = nums[i]
            i = i - 1
        }
        if i < 0 {
            nums.sort()
            return
        }
        var x = nums[i]
        nums[(i+1)...].sort()
        var left = i + 1
        var right = nums.index(before: nums.endIndex)
        
        while right > (left + 1) {
            var m = left + (right - left) / 2
            if nums[m] <= x {
                left = m
            } else {
                right = m
            }
        }
      
        var ind = (x < nums[left]) ? left : right
        nums[i] = nums[ind]
        nums[ind] = x
    }
}

---

32ms

class Solution {
    func nextPermutation(_ nums: inout [Int]) {
        var toSwap = -1
        
        if nums.count <= 1 {
            return
        }
        
        for index in (1...nums.count - 1).reversed() {
            if nums[index] > nums[index - 1] {
                toSwap = index - 1
                break
            }
        }
        
        if toSwap != -1 {
            var min = Int.max, swapIndex = toSwap
            for index in (toSwap + 1...nums.count - 1).reversed() {
                if nums[index] < min && nums[index] > nums[toSwap] {
                    swapIndex = index
                    min = nums[index]
                }
            }
            nums.swapAt(toSwap, swapIndex)
            nums[toSwap + 1...nums.count - 1].reverse()
        } else {
            nums.reverse()
        }
    }
}

---

36ms

class Solution {
    func nextPermutation(_ nums: inout [Int]) {
        //no arrangement possible
        if(nums.sorted().reversed() == nums){
            nums = nums.sorted()
            return
        }
        let n = nums.count
        var position = -1
        for i in (0 ..< n - 1).reversed() {
            print(i)
            if nums[i] < nums[i + 1] {
                position = i
                break
            }
        }
        var indexToSwap = -1
        var diff = Int.max
        
        
        for j in (position+1 ..< n).reversed() {
            let d =  nums[j] - nums[position]
                if d < diff && d > 0 {
                    indexToSwap = j
                    diff = d
                }
            }
        nums.swapAt(position, indexToSwap)
        nums = nums[0 ... position] + nums[position + 1 ... nums.count - 1].reversed()
        
        
        
        
        
    }
}

---

44ms

class Solution {
    func nextPermutation(_ nums: inout [Int]) {
        if nums.count < 2 {
            return
        }
        
        var max = nums[nums.count - 1]
        var i = nums.count - 2
        while i >= 0 {
            if nums[i] < max {
                break
            }
            max = nums[i]
            i = i - 1
        }
        if i < 0 {
            nums.sort()
            return
        }
        var x = nums[i]
        nums[(i+1)...].sort()
        var left = i + 1
        var right = nums.index(before: nums.endIndex)
        
        while right > left {
            var m = left + (right - left) / 2
            if nums[m] <= x {
                if nums[m+1] > x {
                    left = m
                    right = m + 1
                    break
                }
                left = m + 1
            } else {
                if nums[m-1] <= x {
                    left = m - 1
                    right = m
                    break
                }
                right = m - 1
            }
        }
      
        nums[i] = nums[right]
        nums[right] = x
    }
}

---

48ms

class Solution {
    func nextPermutation(_ nums: inout [Int]) {
        var left = 0
        var right = nums.count - 1
        for i in (0 ..< nums.count - 1).reversed() {
            if nums[i] >= nums[i + 1] {
                continue
            } else {
                left = i
                break
            }
        }
        
        var nextIndex = 0
        for i in (0 ..< nums.count).reversed() {
            if nums[i] > nums[left] {
                nextIndex = i 
                break
            }
        }

        nums.swapAt(left, nextIndex)
        if nextIndex != 0 {
            left += 1
        }
        while left < right {
            nums.swapAt(left, right)
            left += 1
            right -= 1
        }
    }
}