[Swift]LeetCode1144. 递减元素使数组呈锯齿状 | Decrease Elements To Make Array Zigzag

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/11297773.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

Given an array nums of integers, a move consists of choosing any element and decreasing it by 1.

An array A is a zigzag array if either:

• Every even-indexed element is greater than adjacent elements, ie. A[0] > A[1] < A[2] > A[3] < A[4] > ...
• OR, every odd-indexed element is greater than adjacent elements, ie. A[0] < A[1] > A[2] < A[3] > A[4] < ...

Return the minimum number of moves to transform the given array nums into a zigzag array. 

Example 1:

Input: nums = [1,2,3]
Output: 2
Explanation: We can decrease 2 to 0 or 3 to 1.

Example 2:

Input: nums = [9,6,1,6,2]
Output: 4 

Constraints:

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 1000

---

给你一个整数数组 nums，每次 操作 会从中选择一个元素并 将该元素的值减少 1。

如果符合下列情况之一，则数组 A 就是 锯齿数组：

• 每个偶数索引对应的元素都大于相邻的元素，即 A[0] > A[1] < A[2] > A[3] < A[4] > ...
• 或者，每个奇数索引对应的元素都大于相邻的元素，即 A[0] < A[1] > A[2] < A[3] > A[4] < ...

返回将数组 nums 转换为锯齿数组所需的最小操作次数。 

示例 1：

输入：nums = [1,2,3]
输出：2
解释：我们可以把 2 递减到 0，或把 3 递减到 1。

示例 2：

输入：nums = [9,6,1,6,2]
输出：4 

提示：

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 1000

---

4ms

class Solution {
    func movesToMakeZigzag(_ nums: [Int]) -> Int {
        if nums.count < 3 {
            return 0
        }
        
        let n = nums.count
        var result1 = 0
        var result2 = 0
        
        var i = 1
        while i < n {
            var minValue = min(nums[i-1], i+1 < n ? nums[i+1] : 1000)
            if minValue <= nums[i] {
                result1 += (nums[i]-minValue+1)
            }
            i += 2
        }
        
        i = 0
        while i < n {
            var minValue = min(i > 0 ? nums[i-1] : 1000, i+1 < n ? nums[i+1] : 1000)
            if minValue <= nums[i] {
                result2 += (nums[i]-minValue+1)
            }
            i += 2
        }
        
        return min(result1, result2)
    }
}

---

8ms

class Solution {
    func movesToMakeZigzag(_ nums: [Int]) -> Int {
        var result = [0, 0]
        let n = nums.count 
        var left = 0
        var right = nums.count - 1
        
        for i in 0..<n {
            left = i > 0 ? nums[i-1] : 1001
            right = i + 1 < n ? nums[i+1] : 1001
            result[i%2] += max(0, nums[i] - min(left, right) + 1)
        }
        return result.min()!
    }
}

---

Runtime: 8 ms

Memory Usage: 20.8 MB

class Solution {
    func movesToMakeZigzag(_ nums: [Int]) -> Int {
        var n:Int = nums.count
        var j:Int = 0
        var s:Int = 0
        var t:Int = 0
        for i in stride(from:0,to:n,by:2)
        {
            j = 0
            if i != 0
            {
                j = max(j,nums[i]-nums[i-1]+1)
            }
            if i + 1 < n
            {
                j = max(j,nums[i]-nums[i+1]+1)
            }
            s += j
        }
        for i in stride(from:1,to:n,by:2)
        {
            j = 0
            if i != 0
            {
                j = max(j,nums[i]-nums[i-1]+1)
            }
            if i + 1 < n
            {
                j = max(j,nums[i]-nums[i+1]+1)
            }
            t += j
        }
        return min(s,t)       
    }
}

---

12ms

class Solution {
    func movesToMakeZigzag(_ nums: [Int]) -> Int {
        var movesA = 0
        var movesB = 0
        var array = nums
        for i in stride(from: 1, to: array.count, by: 2) {
            while array[i - 1] <= array[i] {
                    movesA += 1
                    array[i] -= 1
                }
            
            guard (i + 1) < array.count else { break }
            
            while array[i + 1] <= array[i] {
                    movesA += 1
                    array[i] -= 1
                }
        }
        array = nums
        for i in stride(from: 1, to: array.count, by: 2) {
            while array[i - 1] >= array[i] {
                    movesB += 1
                    array[i - 1] -= 1
                }
            
            guard (i + 1) < array.count else { break }
            
            while array[i + 1] >= array[i] {
                    movesB += 1
                    array[i + 1] -= 1
                }
        }
        
        return movesA < movesB ? movesA : movesB
    }
}