[Swift]LeetCode1043. 分隔數組以獲得最大和 | Partition Array for Maximum Sum

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Given an integer array A, you partition the array into (contiguous) subarrays of length at most K.  After partitioning, each subarray has their values changed to become the maximum value of that subarray.

Return the largest sum of the given array after partitioning.

Example 1:

Input: A = [1,15,7,9,2,5,10], K = 3 Output: 84 Explanation: A becomes [15,15,15,9,10,10,10]

Note:

1. 1 <= K <= A.length <= 500
2. 0 <= A[i] <= 10^6

---

給出整數數組 A，將該數組分隔爲長度最多爲 K 的幾個（連續）子數組。分隔完成後，每一個子數組的中的值都會變爲該子數組中的最大值。

返回給定數組完成分隔後的最大和。

示例：

輸入：A = [1,15,7,9,2,5,10], K = 3
輸出：84
解釋：A 變爲 [15,15,15,9,10,10,10]

提示：

1. 1 <= K <= A.length <= 500
2. 0 <= A[i] <= 10^6

---

32ms

 1 class Solution {
 2     func maxSumAfterPartitioning(_ A: [Int], _ K: Int) -> Int {
 3         guard A.count > 0 else { return 0 }
 4         
 5         var dp = Array(repeating: 0, count: A.count)
 6         var m = 0
 7         for i in 1...A.count {
 8             let j = A.count - i
 9             m = max(m, A[j])
10             if i <= K {
11                 dp[j] = i * m
12             } else {
13                 var result = 0
14                 var localM = 0
15                 for this in j..<(j + K) {
16                     localM = max(localM, A[this])
17                     result = max(result, localM * (this - j + 1) + dp[this + 1])
18                 }
19                 dp[j] = result
20             }
21         }
22         
23         return dp[0]
24     }
25 }

---

Runtime: 56 ms

Memory Usage: 20.9 MB

 1 class Solution {
 2     func maxSumAfterPartitioning(_ A: [Int], _ K: Int) -> Int {
 3         let n:Int = A.count
 4         var dp:[Int] = [Int](repeating:0,count:n + 1)
 5         for i in 1...n
 6         {
 7             var maxs:Int = 0
 8             var j:Int = 1
 9             while(j <= K && i - j >= 0)
10             {
11                 maxs = max(maxs, A[i - j])
12                 dp[i] = max(dp[i], dp[i - j] + maxs * j)
13                 j += 1
14             }
15         }
16         return dp[n]
17     }
18 }

---

64ms 

 1 class Solution {
 2     func maxSumAfterPartitioning(_ A: [Int], _ K: Int) -> Int {
 3         let N = A.count 
 4         var dp = [Int](repeating: 0, count: N)
 5         for i in 0..<N {
 6             var curMax = 0 
 7             for k in 1...K where i - k + 1 >= 0 {
 8                 curMax = max(curMax, A[i - k + 1])
 9                 dp[i] = max(dp[i], (i >= k ? dp[i-k] : 0) + curMax * k)
10             }
11         }
12         return dp[N-1]
13     }
14 }

---

144ms

 1 class Solution {
 2 
 3     var memo = [Int: Int]()
 4     func maxSumAfterPartitioning(_ A: [Int], _ K: Int) -> Int {
 5         var sum_memo = [[Int]](repeating: [Int](repeating: 0, count: K), count: A.count)
 6         for i in A.indices {
 7             var curMax = A[i]
 8             for j in 0..<K {
 9                 if i+j >= A.count { continue }
10                 curMax = max(curMax, A[i+j])
11                 sum_memo[i][j] = curMax * (j+1)
12             }
13         }
14         return maxPartitioningSum(A, K, 0, A.count-1, sum_memo)
15     }
16 
17     func maxPartitioningSum(_ A: [Int], _ K: Int, _ start: Int, _ end: Int, _ summemo: [[Int]]) -> Int {
18         if start > end {
19             return 0
20         }
21 
22         if end - start < K {
23             return summemo[start][end-start]
24         }
25         var ans = 0
26         for i in start..<(start+K) {
27             let index = A.count*(i+1) + end
28             if memo[index] == nil {
29                 memo[index] = maxPartitioningSum(A, K, i+1, end, summemo)
30             }
31             ans = max(ans,summemo[start][i-start] + memo[index]!)
32         }
33         return ans
34     }
35 }

---

152ms

 1 class Solution {
 2     func maxSumAfterPartitioning(_ A: [Int], _ K: Int) -> Int {
 3         var dp = [Int](repeating: 0, count: A.count + 1)
 4         dp[0] = 0
 5         
 6         for i in 1...A.endIndex {
 7             var m = A[i - 1]
 8             for j in stride(from: i, through: max(i - K + 1, 1), by: -1) {
 9             // for j in max(i-K+1, 1)...i {
10                 m = max(m, A[j - 1])
11                 dp[i] = max(dp[i], (m * (i - j + 1)) + dp[j - 1])
12             }
13         }
14         return dp.last!
15     }
16 }