[Swift]LeetCode813. 最大平均值和的分组 | Largest Sum of Averages

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We partition a row of numbers A into at most K adjacent (non-empty) groups, then our score is the sum of the average of each group. What is the largest score we can achieve?

Note that our partition must use every number in A, and that scores are not necessarily integers.

Example:
Input: 
A = [9,1,2,3,9]
K = 3
Output: 20
Explanation: 
The best choice is to partition A into [9], [1, 2, 3], [9]. The answer is 9 + (1 + 2 + 3) / 3 + 9 = 20.
We could have also partitioned A into [9, 1], [2], [3, 9], for example.
That partition would lead to a score of 5 + 2 + 6 = 13, which is worse.

Note:

• 1 <= A.length <= 100.
• 1 <= A[i] <= 10000.
• 1 <= K <= A.length.
• Answers within 10^-6 of the correct answer will be accepted as correct.

---

咱们将给定的数组 A 分红 K 个相邻的非空子数组 ，咱们的分数由每一个子数组内的平均值的总和构成。计算咱们所能获得的最大分数是多少。

注意咱们必须使用 A 数组中的每个数进行分组，而且分数不必定须要是整数。

示例:
输入: 
A = [9,1,2,3,9]
K = 3
输出: 20
解释: 
A 的最优分组是[9], [1, 2, 3], [9]. 获得的分数是 9 + (1 + 2 + 3) / 3 + 9 = 20.
咱们也能够把 A 分红[9, 1], [2], [3, 9].
这样的分组获得的分数为 5 + 2 + 6 = 13, 但不是最大值.

说明:

• 1 <= A.length <= 100.
• 1 <= A[i] <= 10000.
• 1 <= K <= A.length.
• 答案偏差在 10^-6 内被视为是正确的。

---

Runtime: 64 ms

Memory Usage: 19.2 MB

 1 class Solution {
 2     func largestSumOfAverages(_ A: [Int], _ K: Int) -> Double {
 3         var n:Int = A.count
 4         var sums:[Double] = [Double](repeating:0,count:n + 1)
 5         var dp:[Double] = [Double](repeating:0,count:n + 1)
 6         for i in 0..<n
 7         {
 8             sums[i + 1] = sums[i] + Double(A[i])
 9         }
10         for i in 0..<n
11         {
12             dp[i] = (sums[n] - sums[i]) / Double(n - i)
13         }
14         for k in 1..<K
15         {
16             for i in 0..<(n - 1)
17             {
18                 for j in (i + 1)..<n
19                 {
20                     dp[i] = max(dp[i], (sums[j] - sums[i]) / Double(j - i) + dp[j])
21                 }
22             }
23         }
24         return dp[0]
25     }
26 }

---

92ms

 1 private struct State: Hashable {
 2     var i: Int
 3     var K: Int
 4 }
 5 
 6 class Solution {
 7     func largestSumOfAverages(_ A: [Int], _ K: Int) -> Double {
 8         var dict = [State: Double]()
 9         return largestSumOfAverages(A, K, 0, &dict)
10     }
11     
12     private func largestSumOfAverages(_ A: [Int], _ K: Int, _ i: Int, _ dict: inout [State: Double]) -> Double {
13         guard i < A.count else { return 0 }
14         if let val = dict[State(i: i, K: K)] {
15             return val
16         }
17         var sum: Double = 0.0
18         var largest: Double = 0.0
19         for j in i..<A.count {
20             sum += Double(A[j])
21             let avg = sum / Double(j - i + 1)
22             if K - 1 > 0 {
23                 largest = max(largest, avg + largestSumOfAverages(A, K - 1, j + 1, &dict))
24             }
25         }
26         largest = max(largest, sum / Double(A.count - i))
27         dict[State(i: i, K: K)] = largest
28         return largest
29     }
30 }

---

100ms

 1 class Solution {    
 2     // memo[k][i] 前 i 个 元素分 k 份产生的最大值
 3     var memo: [[Double]]!
 4     var sum: [Double]!
 5     
 6     func largestSumOfAverages(_ A: [Int], _ K: Int) -> Double {
 7         let n = A.count
 8         if n == 0 {
 9             return 0
10         }
11         memo = Array(repeating: Array(repeating: 0.0, count: n+1), count: K+1)
12         sum = Array(repeating: 0, count: n+1)
13         for i in 1...n {
14             sum[i] = sum[i-1] + Double(A[i-1])
15         }
16         return LSA(A, n, K)
17     }
18     
19     func LSA(_ A: [Int], _ n: Int, _ k: Int) -> Double {
20         if memo[k][n] > 0 {
21             return memo[k][n]
22         }
23         if k == 1 {
24             return sum[n] / Double(n)
25         }
26         for i in k-1..<n {
27             // 老是要分红最后一个和前面几个的, 只是不知道那里是最后一个, 有点像走一步两步的那个思路
28             memo[k][n] = max(memo[k][n], LSA(A, i, k-1) + (sum[n] - sum[i]) / Double(n - i))
29         }
30         return memo[k][n]
31     }
32 }

---

188ms

 1 class Solution {
 2     func largestSumOfAverages(_ A: [Int], _ K: Int) -> Double {
 3         var cache = [State: Double]()
 4         return largestSum(A, K, 0, &cache)
 5     }
 6     
 7     func largestSum(_ A: [Int], _ K: Int, _ index: Int, _ cache: inout [State: Double]) -> Double {
 8         let count = A.count
 9         guard count > index else { return 0 }
10         var sum: Double = 0
11         for i in index..<count { sum += Double(A[i]) }
12             
13         guard K > 1 else {
14             return sum / Double(count - index)
15         }
16         let state = State(x: K, y: index)
17         if let val = cache[state] { return val }
18         var mx: Double = 0
19         sum = 0
20         var totalElements = 0
21         for i in index..<count {
22             totalElements += 1
23             sum += Double(A[i])
24             mx = max(mx, sum / Double(totalElements) + largestSum(A, K - 1, i + 1, &cache))
25         }
26         cache[state] = mx
27         return mx
28     }
29 }
30 
31 struct State: Hashable {
32     var x: Int
33     var y: Int
34 }