[Swift]LeetCode665. 非递减数列 | Non-decreasing Array

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Given an array with n integers, your task is to check if it could become non-decreasing by modifying at most 1 element.

We define an array is non-decreasing if array[i] <= array[i + 1] holds for every i (1 <= i < n).

Example 1:

Input: [4,2,3]
Output: True
Explanation: You could modify the first 4 to 1 to get a non-decreasing array. 

Example 2:

Input: [4,2,1]
Output: False
Explanation: You can't get a non-decreasing array by modify at most one element. 

Note: The n belongs to [1, 10,000].

---

给定一个长度为 n 的整数数组，你的任务是判断在最多改变 1 个元素的情况下，该数组能否变成一个非递减数列。

我们是这样定义一个非递减数列的： 对于数组中所有的 i (1 <= i < n)，满足 array[i] <= array[i + 1]。

示例 1:

输入: [4,2,3]
输出: True
解释: 你可以通过把第一个4变成1来使得它成为一个非递减数列。

示例 2:

输入: [4,2,1]
输出: False
解释: 你不能在只改变一个元素的情况下将其变为非递减数列。

说明:  n 的范围为 [1, 10,000]。

---

44ms

class Solution {
    func checkPossibility(_ nums: [Int]) -> Bool {
        var i = 0
        var j = nums.count - 1
        while i < j && nums[i] <= nums[i+1] {
            i += 1
        }
        while i < j && nums[j] >= nums[j-1] {
            j -= 1
        }
        let head = (i == 0) ? Int.min : nums[i-1]
        let next = (j == nums.count - 1) ? Int.max : nums[j+1]
        
        if j - i <= 1 && (head < nums[j] || nums[i] < next) {
            return true
        } else {
            return false
        }
    }
}

---

188ms

class Solution {
    func checkPossibility(_ nums: [Int]) -> Bool {
        
        if nums.count < 3 {
            return true
        }
        
        var i = 0
        
        while i <= nums.count - 2, nums[i] <= nums[i + 1] {
            i += 1
        }
        
        var j = nums.count - 1
        
        if i >= j - 1 {
            return true
        }
        
        while 1 <= j, nums[j - 1] <= nums[j] {
            j -= 1
        }
        
        if j <= 1  {
            return true
        }
        
        return ((j - i == 2) && (nums[i] <= nums[j])) 
               || 
               ((j - i == 1) 
                   && 
                   ((0 < i) && (nums[i - 1] <= nums[j])
                   || ((j < nums.count - 1) && (nums[i] <= nums[j + 1]))))
    }
}

---

Runtime: 192 ms

Memory Usage: 19.5 MB

class Solution {
    func checkPossibility(_ nums: [Int]) -> Bool {
        var nums = nums
        var cnt:Int = 1
        var n:Int = nums.count
        for i in 1..<n
        {
            if nums[i] < nums[i - 1]
            {
                if cnt == 0 {return false}
                if i == 1 || nums[i] >= nums[i - 2]
                {
                    nums[i - 1] = nums[i]
                }
                else
                {
                    nums[i] = nums[i - 1]
                }
                cnt -= 1
            }
        }
        return true
    }
}

---

196ms

class Solution {
    func checkPossibility(_ nums: [Int]) -> Bool {
            var c = 0
            var nums = [-Int.max] + nums
            var low = nums[0]
        
        for (i, n) in nums.enumerated().dropFirst() {
            if n < nums[i-1] {
                if n >= low {
                    nums[i-1] = low
                } else {
                    nums[i] = nums[i-1]
                }
                c += 1
                if c > 1 {
                    return false
                }
            }
            if i > 1 {
                low = nums[i-1]
            }
        }
        
        return true
    }
}

---

200ms

class Solution {
    func checkPossibility(_ nums: [Int]) -> Bool {
        var newNums = nums
        if newNums.count <= 2 {
            return true
        }
        
        var i = 1
        var j = 2
        var noOfChange = 0
        
        if (newNums[i] < newNums[i-1]) {
            newNums[i-1] = newNums[i]
            noOfChange = noOfChange + 1
        }
        
        while j <= newNums.count - 1 {
            if (newNums[i] > newNums[j]) {
                //if [j] > [i-1], change value of [i] -> [i-1] 
                if (newNums[j] > newNums[i-1]) {
                    newNums[i] = newNums[i-1]
                } 
                //else, change [j] -> [i]
                else { 
                    newNums[j] = newNums[i]
                }
                
                noOfChange = noOfChange + 1
            }
            j = j + 1
            i = i + 1
        }
        
        return noOfChange <= 1
    }
}

---

208ms

class Solution {
    func checkPossibility(_ nums: [Int]) -> Bool {
        
        if nums.count < 3 {
            return true
        }
        
        var index = -1
        
        for i in 0 ..< (nums.count - 1) {
            if nums[i] > nums[i+1] {
                if index != -1 { return false }
                index = i
            }
        }
        
        return ((index == -1)
            || (index == 0)
            || (index == nums.count - 2)
            || (nums[index+1] - nums[index-1] > 0)
            || (nums[index+2] - nums[index] > 0))
    }
}

---

284ms

class Solution {
    func checkPossibility(_ nums: [Int]) -> Bool {
        guard nums.count > 1 && nums.count <= 10000 else {
            return nums.count == 1
        }
        var array = nums
        var modified = false
        for i in 0 ..< array.count - 1 {
            if array[i] > array[i + 1] {
                if modified {
                    return false
                }
                if i > 0 && array[i - 1] > array[i + 1] {
                    array[i + 1] = array[i]
                }
                modified = true
            }
        }
        return true
    }
}