[Swift]LeetCode1144. 遞減元素使數組呈鋸齒狀 | Decrease Elements To Make Array Zigzag

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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

---

給你一個整數數組 nums，每次 操做 會從中選擇一個元素並 將該元素的值減小 1。

若是符合下列狀況之一，則數組 A 就是 鋸齒數組：

• 每一個偶數索引對應的元素都大於相鄰的元素，即 A[0] > A[1] < A[2] > A[3] < A[4] > ...
• 或者，每一個奇數索引對應的元素都大於相鄰的元素，即 A[0] < A[1] > A[2] < A[3] > A[4] < ...

返回將數組 nums 轉換爲鋸齒數組所需的最小操做次數。 

示例 1：

輸入：nums = [1,2,3]
輸出：2
解釋：咱們能夠把 2 遞減到 0，或把 3 遞減到 1。

示例 2：

輸入：nums = [9,6,1,6,2]
輸出：4 

提示：

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

---

4ms

 1 class Solution {
 2     func movesToMakeZigzag(_ nums: [Int]) -> Int {
 3         if nums.count < 3 {
 4             return 0
 5         }
 6         
 7         let n = nums.count
 8         var result1 = 0
 9         var result2 = 0
10         
11         var i = 1
12         while i < n {
13             var minValue = min(nums[i-1], i+1 < n ? nums[i+1] : 1000)
14             if minValue <= nums[i] {
15                 result1 += (nums[i]-minValue+1)
16             }
17             i += 2
18         }
19         
20         i = 0
21         while i < n {
22             var minValue = min(i > 0 ? nums[i-1] : 1000, i+1 < n ? nums[i+1] : 1000)
23             if minValue <= nums[i] {
24                 result2 += (nums[i]-minValue+1)
25             }
26             i += 2
27         }
28         
29         return min(result1, result2)
30     }
31 }

---

8ms

 1 class Solution {
 2     func movesToMakeZigzag(_ nums: [Int]) -> Int {
 3         var result = [0, 0]
 4         let n = nums.count 
 5         var left = 0
 6         var right = nums.count - 1
 7         
 8         for i in 0..<n {
 9             left = i > 0 ? nums[i-1] : 1001
10             right = i + 1 < n ? nums[i+1] : 1001
11             result[i%2] += max(0, nums[i] - min(left, right) + 1)
12         }
13         return result.min()!
14     }
15 }

---

Runtime: 8 ms

Memory Usage: 20.8 MB

 1 class Solution {
 2     func movesToMakeZigzag(_ nums: [Int]) -> Int {
 3         var n:Int = nums.count
 4         var j:Int = 0
 5         var s:Int = 0
 6         var t:Int = 0
 7         for i in stride(from:0,to:n,by:2)
 8         {
 9             j = 0
10             if i != 0
11             {
12                 j = max(j,nums[i]-nums[i-1]+1)
13             }
14             if i + 1 < n
15             {
16                 j = max(j,nums[i]-nums[i+1]+1)
17             }
18             s += j
19         }
20         for i in stride(from:1,to:n,by:2)
21         {
22             j = 0
23             if i != 0
24             {
25                 j = max(j,nums[i]-nums[i-1]+1)
26             }
27             if i + 1 < n
28             {
29                 j = max(j,nums[i]-nums[i+1]+1)
30             }
31             t += j
32         }
33         return min(s,t)       
34     }
35 }

---

12ms

 

 1 class Solution {
 2     func movesToMakeZigzag(_ nums: [Int]) -> Int {
 3         var movesA = 0
 4         var movesB = 0
 5         var array = nums
 6         for i in stride(from: 1, to: array.count, by: 2) {
 7             while array[i - 1] <= array[i] {
 8                     movesA += 1
 9                     array[i] -= 1
10                 }
11             
12             guard (i + 1) < array.count else { break }
13             
14             while array[i + 1] <= array[i] {
15                     movesA += 1
16                     array[i] -= 1
17                 }
18         }
19         array = nums
20         for i in stride(from: 1, to: array.count, by: 2) {
21             while array[i - 1] >= array[i] {
22                     movesB += 1
23                     array[i - 1] -= 1
24                 }
25             
26             guard (i + 1) < array.count else { break }
27             
28             while array[i + 1] >= array[i] {
29                     movesB += 1
30                     array[i + 1] -= 1
31                 }
32         }
33         
34         return movesA < movesB ? movesA : movesB
35     }
36 }