[Swift]LeetCode1031. 兩個非重疊子數組的最大和 | Maximum Sum of Two Non-Overlapping Subarrays

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公衆號：山青詠芝（shanqingyongzhi）
➤博客園地址：山青詠芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址： http://www.javashuo.com/article/p-wuzyohsq-md.html 
➤若是連接不是山青詠芝的博客園地址，則多是爬取做者的文章。
➤原文已修改更新！強烈建議點擊原文地址閱讀！支持做者！支持原創！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

Given an array A of non-negative integers, return the maximum sum of elements in two non-overlapping (contiguous) subarrays, which have lengths L and M.  (For clarification, the L-length subarray could occur before or after the M-length subarray.)

Formally, return the largest V for which V = (A[i] + A[i+1] + ... + A[i+L-1]) + (A[j] + A[j+1] + ... + A[j+M-1]) and either:

• 0 <= i < i + L - 1 < j < j + M - 1 < A.length, or
• 0 <= j < j + M - 1 < i < i + L - 1 < A.length.

Example 1:

Input: A = [0,6,5,2,2,5,1,9,4], L = 1, M = 2 Output: 20 Explanation: One choice of subarrays is [9] with length 1, and [6,5] with length 2. 

Example 2:

Input: A = [3,8,1,3,2,1,8,9,0], L = 3, M = 2 Output: 29 Explanation: One choice of subarrays is [3,8,1] with length 3, and [8,9] with length 2. 

Example 3:

Input: A = [2,1,5,6,0,9,5,0,3,8], L = 4, M = 3 Output: 31 Explanation: One choice of subarrays is [5,6,0,9] with length 4, and [3,8] with length 3.

Note:

1. L >= 1
2. M >= 1
3. L + M <= A.length <= 1000
4. 0 <= A[i] <= 1000

---

給出非負整數數組 A ，返回兩個非重疊（連續）子數組中元素的最大和，子數組的長度分別爲 L 和 M。（這裏須要澄清的是，長爲 L 的子數組能夠出如今長爲 M 的子數組以前或以後。）

從形式上看，返回最大的 V，而 V = (A[i] + A[i+1] + ... + A[i+L-1]) + (A[j] + A[j+1] + ... + A[j+M-1]) 並知足下列條件之一：

• 0 <= i < i + L - 1 < j < j + M - 1 < A.length, 或
• 0 <= j < j + M - 1 < i < i + L - 1 < A.length.

示例 1：

輸入：A = [0,6,5,2,2,5,1,9,4], L = 1, M = 2
輸出：20
解釋：子數組的一種選擇中，[9] 長度爲 1，[6,5] 長度爲 2。

示例 2：

輸入：A = [3,8,1,3,2,1,8,9,0], L = 3, M = 2
輸出：29
解釋：子數組的一種選擇中，[3,8,1] 長度爲 3，[8,9] 長度爲 2。

示例 3：

輸入：A = [2,1,5,6,0,9,5,0,3,8], L = 4, M = 3
輸出：31
解釋：子數組的一種選擇中，[5,6,0,9] 長度爲 4，[0,3,8] 長度爲 3。

提示：

1. L >= 1
2. M >= 1
3. L + M <= A.length <= 1000
4. 0 <= A[i] <= 1000

---

Runtime: 24 ms

Memory Usage: 19.1 MB

 1 class Solution {
 2     let N:Int = 1010
 3     var prefix:[Int] = [Int](repeating:0,count:1010)
 4     func maxSumTwoNoOverlap(_ A: [Int], _ L: Int, _ M: Int) -> Int {
 5         var n:Int = A.count
 6         for i in 1...n
 7         {
 8             prefix[i] = prefix[i - 1] + A[i - 1]
 9         }
10         var ans:Int = 0
11         var best_l:Int = 0
12         for i in (L + M)...n
13         {
14             best_l = max(best_l, prefix[i - M] - prefix[i - M - L])
15             ans = max(ans, best_l + prefix[i] - prefix[i - M])
16         }
17         var best_m:Int = 0
18         for i in (L + M)...n
19         {
20             best_m = max(best_m, prefix[i - L] - prefix[i - M - L])
21             ans = max(ans, best_m + prefix[i] - prefix[i - L])
22         }
23         return ans        
24     }
25 }

---

24ms 

 1 class Solution {
 2     func maxSumTwoNoOverlap(_ A: [Int], _ L: Int, _ M: Int) -> Int {
 3         var leftMax = [Int](repeating: 0, count: A.count)
 4         var curMax = 0, curSum = 0
 5         for i in A.indices {
 6             curSum += i < M ?  A[i] : A[i] - A[i-M]
 7             curMax = max(curMax, curSum)
 8             leftMax[i] = curMax
 9         }
10 
11         var rightMax = [Int](repeating: 0, count: A.count+1)
12         var i = A.count - 1, ans = 0
13         curMax = 0; curSum = 0
14         while i >= 0 {
15             curSum += (i + M > A.count - 1) ? A[i] : A[i] - A[i+M]
16             curMax = max(curMax, curSum)
17             rightMax[i] = curMax
18             i -= 1
19         }
20 
21         curSum = 0
22         for i in A.indices {
23             curSum += i < L ?  A[i] : A[i] - A[i-L]
24             let leftMax =  i >= L ? leftMax[i-L] : 0
25             ans = max(ans, max(leftMax, rightMax[i+1]) + curSum)
26         }
27         return ans
28     }
29 }

---

84ms

 1 class Solution {
 2     func maxSumTwoNoOverlap(_ A: [Int], _ L: Int, _ M: Int) -> Int {
 3         var sumLs = [Int](repeating: 0, count: A.count)
 4         var sumMs = [Int](repeating: 0, count: A.count)
 5         var forwardMaxLs = [Int](repeating: 0, count: A.count)
 6         var backwardMaxLs = [Int](repeating: 0, count: A.count)
 7         var forwardMaxMs = [Int](repeating: 0, count: A.count)
 8         var backwardMaxMs = [Int](repeating: 0, count: A.count)
 9 
10         var sumL = 0
11         var sumM = 0
12         for i in 0..<A.count {
13             sumL += A[i]
14             if i >= L - 1 {
15                 if i >= L {
16                     sumL -= A[i - L]
17                 }
18                 sumLs[i] = sumL
19             }
20             sumM += A[i]
21             if i >= M - 1 {
22                 if i >= M {
23                     sumM -= A[i - M]
24                 }
25                 sumMs[i] = sumM
26             }
27         }
28         var maxSumL = Int.min
29         var maxSumM = Int.min
30         for i in 0..<A.count {
31             maxSumL = max(maxSumL, sumLs[i])
32             forwardMaxLs[i] = maxSumL
33             maxSumM = max(maxSumM, sumMs[i])
34             forwardMaxMs[i] = maxSumM
35         }
36         maxSumL = Int.min
37         maxSumM = Int.min
38         for i in (0..<A.count).reversed() {
39             maxSumL = max(maxSumL, sumLs[i])
40             backwardMaxLs[i] = maxSumL
41             maxSumM = max(maxSumM, sumMs[i])
42             backwardMaxMs[i] = maxSumM
43         }
44         var maxSum = Int.min
45         for l in 0..<A.count {
46             var m = A.count - 1
47             while l <= m - M {
48                 maxSum = max(maxSum, forwardMaxLs[l] + backwardMaxMs[m])
49                 m -= 1
50             }
51         }
52         for m in 0..<A.count {
53             var l = A.count - 1
54             while m <= l - L {
55                 maxSum = max(maxSum, backwardMaxLs[l] + forwardMaxMs[m])
56                 l -= 1
57             }
58         }
59         return maxSum
60     }
61 }

---

108ms

 1 class Solution {
 2     func maxSumTwoNoOverlap(_ A: [Int], _ L: Int, _ M: Int) -> Int {
 3         var prefixSums: [Int] = [0]
 4         var sum = 0
 5         for num in A {
 6             sum += num
 7             prefixSums.append(sum)
 8         }
 9         var res = 0
10         for lRight in L..<prefixSums.count {
11             let lSum = prefixSums[lRight] - prefixSums[lRight - L]
12             for mRight in stride(from: lRight + M, to: prefixSums.count, by: 1) {
13                 let mSum = prefixSums[mRight] - prefixSums[mRight - M]
14                 let total = mSum + lSum
15                 res = max(res, total)
16             }
17             
18             for mRight in stride(from: M, to: lRight - L, by: 1) {
19                 let mSum = prefixSums[mRight] - prefixSums[mRight - M]
20                 let total = mSum + lSum
21                 res = max(res, total)
22             }
23         }
24         return res
25     }
26 }