[Swift]LeetCode1043.分隔数组以得到最大和|PartitionArrayforMaximumSum

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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 = 3
Output: 84
Explanation: A becomes [15,15,15,9,10,10,10]

Note:

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

给出整数数组 A，将该数组分隔为长度最多为 K 的几个（连续）子数组。分隔完成后，每个子数组的中的值都会变为该子数组中的最大值。

返回给定数组完成分隔后的最大和。

示例：

输入：A = [1,15,7,9,2,5,10], K = 3
输出：84
解释：A 变为 [15,15,15,9,10,10,10]

提示：

1 <= K <= A.length <= 500
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 }