[Swift]LeetCode1186. 刪除一次獲得子數組最大和 | Maximum Subarray Sum with One Deletion

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Given an array of integers, return the maximum sum for a non-empty subarray (contiguous elements) with at most one element deletion. In other words, you want to choose a subarray and optionally delete one element from it so that there is still at least one element left and the sum of the remaining elements is maximum possible.

Note that the subarray needs to be non-empty after deleting one element.

 

Example 1:

Input: arr = [1,-2,0,3]
Output: 4
Explanation: Because we can choose [1, -2, 0, 3] and drop -2, thus the subarray [1, 0, 3] becomes the maximum value.

Example 2:

Input: arr = [1,-2,-2,3]
Output: 3
Explanation: We just choose [3] and it's the maximum sum.

Example 3:

Input: arr = [-1,-1,-1,-1]
Output: -1
Explanation: The final subarray needs to be non-empty. You can't choose [-1] and delete -1 from it, then get an empty subarray to make the sum equals to 0.

 

Constraints:

• 1 <= arr.length <= 10^5
• -10^4 <= arr[i] <= 10^4

---

給你一個整數數組，返回它的某個 非空 子數組（連續元素）在執行一次可選的刪除操做後，所能獲得的最大元素總和。

換句話說，你能夠從原數組中選出一個子數組，並能夠決定要不要從中刪除一個元素（只能刪一次哦），（刪除後）子數組中至少應當有一個元素，而後該子數組（剩下）的元素總和是全部子數組之中最大的。

注意，刪除一個元素後，子數組 不能爲空。

請看示例：

示例 1：

輸入：arr = [1,-2,0,3]
輸出：4
解釋：咱們能夠選出 [1, -2, 0, 3]，而後刪掉 -2，這樣獲得 [1, 0, 3]，和最大。

示例 2：

輸入：arr = [1,-2,-2,3]
輸出：3
解釋：咱們直接選出 [3]，這就是最大和。

示例 3：

輸入：arr = [-1,-1,-1,-1]
輸出：-1
解釋：最後獲得的子數組不能爲空，因此咱們不能選擇 [-1] 並從中刪去 -1 來獲得 0。
     咱們應該直接選擇 [-1]，或者選擇 [-1, -1] 再從中刪去一個 -1。

 

提示：

• 1 <= arr.length <= 10^5
• -10^4 <= arr[i] <= 10^4

---

Runtime: 236 ms

Memory Usage: 24.5 MB

 1 class Solution {
 2     func maximumSum(_ arr: [Int]) -> Int {
 3         let n:Int = arr.count
 4         if n == 1
 5         {
 6             return arr.first!
 7         }
 8         var leftMaxSum:[Int] = [Int](repeating:0,count:n)
 9         for i in 1..<n
10         {
11             leftMaxSum[i] = max(arr[i - 1], leftMaxSum[i - 1] + arr[i - 1])
12         }
13         var rightMaxSum:[Int] = [Int](repeating:0,count:n)
14         for i in stride(from:n - 2,through:0,by:-1)
15         {
16             rightMaxSum[i] = max(arr[i + 1], rightMaxSum[i + 1] + arr[i + 1])
17         }
18         var maxNum:Int = max(leftMaxSum[n - 1], rightMaxSum[0])
19         for i in 1..<(n - 1)
20         {
21             maxNum = max(maxNum, leftMaxSum[i])
22             maxNum = max(maxNum, rightMaxSum[i])
23             maxNum = max(maxNum, leftMaxSum[i] + rightMaxSum[i])
24         }
25         return maxNum
26     }
27 }

---

 244ms

 1 class Solution {
 2     func maximumSum(_ arr: [Int]) -> Int {
 3         if arr.isEmpty { return 0 }
 4     if arr.count == 1 { return arr[0] }
 5         var forward = Array(repeating: 0, count: arr.count)
 6     var backward = Array(repeating: 0, count: arr.count)
 7     var currMax = arr[0], overMax = arr[0], answer = 0
 8     forward[0] = arr[0]
 9     for i in 1..<arr.count {
10         currMax = max(arr[i], currMax + arr[i])
11         overMax = max(overMax, currMax)
12         forward[i] = currMax
13     }
14     currMax = arr[arr.count-1]
15     overMax = arr[arr.count-1]
16     backward[arr.count-1] = arr[arr.count-1]
17     for i in (0..<arr.count-1).reversed() {
18         currMax = max(arr[i], currMax + arr[i])
19         overMax = max(overMax, currMax)
20         backward[i] = currMax
21     }
22     answer = overMax
23     for i in 1..<arr.count-1 {
24         answer = max(answer, forward[i-1]+backward[i+1])
25     }
26     return answer
27     }
28 }

---

248ms

 1 class Solution {
 2     func maximumSum(_ arr: [Int]) -> Int {
 3                 
 4         if arr.count == 0 {
 5             return -1
 6         }
 7         if arr.count == 1 {
 8             return arr[0]
 9         }
10 
11         var dp = [Int](repeating: 0, count: arr.count)
12         dp[0] = arr[0]
13         var max1 = dp[0]
14         for i in 1..<arr.count {
15             if dp[i-1] >= 0 {
16                 dp[i] = arr[i] + dp[i-1]
17             } else {
18                 dp[i] = arr[i]
19             }
20             max1 = max(max1, dp[i])
21         }
22 
23         var dp2 = [Int](repeating: 0, count: arr.count)
24         dp2[arr.count-1] = arr[arr.count-1]
25         var max2 = dp2[arr.count-1]
26         for i in stride(from: arr.count-2, to: -1, by: -1) {
27             if dp2[i+1] > 0 {
28                 dp2[i] = arr[i] + dp2[i+1]
29             } else {
30                 dp2[i] = arr[i]
31             }
32             max2 = max(dp2[i], max2)
33         }
34         var max3 = max(max1, max2)
35         
36         for i in 1..<arr.count-1 {
37             max3 = max(dp[i-1]+dp2[i+1], max3)
38         }
39         return max3
40     }
41 }