[Swift]LeetCode956. 最高的广告牌 | Tallest Billboard

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You are installing a billboard and want it to have the largest height.  The billboard will have two steel supports, one on each side.  Each steel support must be an equal height.

You have a collection of rods which can be welded together.  For example, if you have rods of lengths 1, 2, and 3, you can weld them together to make a support of length 6.

Return the largest possible height of your billboard installation.  If you cannot support the billboard, return 0.

Example 1:

Input: [1,2,3,6]
Output: 6
Explanation: We have two disjoint subsets {1,2,3} and {6}, which have the same sum = 6.

Example 2:

Input: [1,2,3,4,5,6]
Output: 10
Explanation: We have two disjoint subsets {2,3,5} and {4,6}, which have the same sum = 10.

Example 3:

Input: [1,2]
Output: 0
Explanation: The billboard cannot be supported, so we return 0.

Note:

1. 0 <= rods.length <= 20
2. 1 <= rods[i] <= 1000
3. The sum of rods is at most 5000.

---

你正在安装一个广告牌，并希望它高度最大。这块广告牌将有两个钢制支架，两边各一个。每个钢支架的高度必须相等。

你有一堆可以焊接在一起的钢筋 rods。举个例子，如果钢筋的长度为 1、2 和 3，则可以将它们焊接在一起形成长度为 6 的支架。

返回广告牌的最大可能安装高度。如果没法安装广告牌，请返回 0。

示例 1：

输入：[1,2,3,6]
输出：6
解释：我们有两个不相交的子集 {1,2,3} 和 {6}，它们具有相同的和 sum = 6。

示例 2：

输入：[1,2,3,4,5,6]
输出：10
解释：我们有两个不相交的子集 {2,3,5} 和 {4,6}，它们具有相同的和 sum = 10。

示例 3：

输入：[1,2]
输出：0
解释：没法安装广告牌，所以返回 0。

提示：

1. 0 <= rods.length <= 20
2. 1 <= rods[i] <= 1000
3. 钢筋的总数最多为 5000 根

---

1392ms

class Solution {
    func tallestBillboard(_ rods: [Int]) -> Int {
        var n:Int = rods.count
        var h:Int = n/2
        var o:Int = 10002
        var ls:[Int] = [Int](repeating:-99999999,count:20005)
        for i in 0..<Int(pow(3, Double(h)))
        {
            var s:Int = 0
            var ass:Int = 0
            var v:Int = i
            for j in 0..<h
            {
                var w:Int = v % 3
                if w == 1
                {
                    
                }
                else if w == 0
                {
                    s += rods[j]
                    ass += rods[j]
                }
                else
                {
                    s -= rods[j]
                    ass += rods[j]
                }
                v /= 3
            }
            ls[s+o] = max(ls[s+o], ass)
        }
        var ret:Int = 0
        for i in 0..<Int(pow(3, Double(n - h)))
        {
            var s:Int = 0
            var ass:Int = 0
            var v:Int = i
            for j in 0..<(n - h)
            {
                var w:Int = v % 3
                if w == 1
                {
                    
                }
                else if w == 0
                {
                    s += rods[j + h]
                    ass += rods[j + h]
                }
                else
                {
                    s -= rods[j + h]
                    ass += rods[j + h]
                }
                v /= 3
            }
            ret = max(ret, (ls[o-s] + ass) / 2)
        }
        return ret
    }
}

---

1480ms

class Solution {
    struct State: Hashable {
        let delta: Int
        let score: Int
    }

    func make(from half: ArraySlice<Int>) -> [Int : Int] {
        var states = Set<State>([.init(delta: 0, score: 0)])
        for i in half {
            let x = Set(states.map { State(delta: $0.delta + i, score: $0.score) })
            let y = Set(states.map { State(delta: $0.delta, score: $0.score + i) })
            states = states.union(x.union(y))
        }

        var delta = [Int : Int]()
        for state in states {
            delta[state.delta - state.score] = max(delta[state.delta - state.score] ?? 0, state.delta)
        }

        return delta
    }

    func tallestBillboard(_ rods: [Int]) -> Int {
        let count = rods.count
        let leftDelta = make(from: rods[..<(count / 2)])
        let rightDelta = make(from: rods[(count / 2)...])

        var answer = 0

        for delta in leftDelta.keys {
            if rightDelta.keys.contains(-delta) {
                answer = max(answer, leftDelta[delta]! + rightDelta[-delta]!)
            }
        }

        return answer
    }
}