[Swift]LeetCode1043. 分隔数组以得到最大和 | Partition Array for Maximum Sum

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/10852127.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎，请直接点击！！！

进入博主App Store主页，下载使用各个作品！！！

注：博主将坚持每月上线一个新app！！！

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

---

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

---

32ms

class Solution {
    func maxSumAfterPartitioning(_ A: [Int], _ K: Int) -> Int {
        guard A.count > 0 else { return 0 }
        
        var dp = Array(repeating: 0, count: A.count)
        var m = 0
        for i in 1...A.count {
            let j = A.count - i
            m = max(m, A[j])
            if i <= K {
                dp[j] = i * m
            } else {
                var result = 0
                var localM = 0
                for this in j..<(j + K) {
                    localM = max(localM, A[this])
                    result = max(result, localM * (this - j + 1) + dp[this + 1])
                }
                dp[j] = result
            }
        }
        
        return dp[0]
    }
}

---

Runtime: 56 ms

Memory Usage: 20.9 MB

class Solution {
    func maxSumAfterPartitioning(_ A: [Int], _ K: Int) -> Int {
        let n:Int = A.count
        var dp:[Int] = [Int](repeating:0,count:n + 1)
        for i in 1...n
        {
            var maxs:Int = 0
            var j:Int = 1
            while(j <= K && i - j >= 0)
            {
                maxs = max(maxs, A[i - j])
                dp[i] = max(dp[i], dp[i - j] + maxs * j)
                j += 1
            }
        }
        return dp[n]
    }
}

---

64ms 

class Solution {
    func maxSumAfterPartitioning(_ A: [Int], _ K: Int) -> Int {
        let N = A.count 
        var dp = [Int](repeating: 0, count: N)
        for i in 0..<N {
            var curMax = 0 
            for k in 1...K where i - k + 1 >= 0 {
                curMax = max(curMax, A[i - k + 1])
                dp[i] = max(dp[i], (i >= k ? dp[i-k] : 0) + curMax * k)
            }
        }
        return dp[N-1]
    }
}

---

144ms

class Solution {

    var memo = [Int: Int]()
    func maxSumAfterPartitioning(_ A: [Int], _ K: Int) -> Int {
        var sum_memo = [[Int]](repeating: [Int](repeating: 0, count: K), count: A.count)
        for i in A.indices {
            var curMax = A[i]
            for j in 0..<K {
                if i+j >= A.count { continue }
                curMax = max(curMax, A[i+j])
                sum_memo[i][j] = curMax * (j+1)
            }
        }
        return maxPartitioningSum(A, K, 0, A.count-1, sum_memo)
    }

    func maxPartitioningSum(_ A: [Int], _ K: Int, _ start: Int, _ end: Int, _ summemo: [[Int]]) -> Int {
        if start > end {
            return 0
        }

        if end - start < K {
            return summemo[start][end-start]
        }
        var ans = 0
        for i in start..<(start+K) {
            let index = A.count*(i+1) + end
            if memo[index] == nil {
                memo[index] = maxPartitioningSum(A, K, i+1, end, summemo)
            }
            ans = max(ans,summemo[start][i-start] + memo[index]!)
        }
        return ans
    }
}

---

152ms

class Solution {
    func maxSumAfterPartitioning(_ A: [Int], _ K: Int) -> Int {
        var dp = [Int](repeating: 0, count: A.count + 1)
        dp[0] = 0
        
        for i in 1...A.endIndex {
            var m = A[i - 1]
            for j in stride(from: i, through: max(i - K + 1, 1), by: -1) {
            // for j in max(i-K+1, 1)...i {
                m = max(m, A[j - 1])
                dp[i] = max(dp[i], (m * (i - j + 1)) + dp[j - 1])
            }
        }
        return dp.last!
    }
}