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.

你正在安裝一個廣告牌，並希望它高度最大。這塊廣告牌將有兩個鋼製支架，兩邊各一個。每個鋼支架的高度必須相等。

你有一堆可以焊接在一起的鋼筋 rods。舉個例子，如果鋼筋的長度為 1、2 和 3，則可以將它們焊接在一起形成長度為 6 的支架。

返回廣告牌的最大可能安裝高度。如果沒法安裝廣告牌，請返回 0。

示例 1：

輸入：[1,2,3,6]
輸出：6
解釋：我們有兩個不相交的子集 {1,2,3} 和 {6}，它們具有相同的和 sum = 6。

示例 2：

輸入：[1,2,3,4,5,6]
輸出：10
解釋：我們有兩個不相交的子集 {2,3,5} 和 {4,6}，它們具有相同的和 sum = 10。

示例 3：

輸入：[1,2]
輸出：0
解釋：沒法安裝廣告牌，所以返回 0。

提示：

1. 0 <= rods.length <= 20
2. 1 <= rods[i] <= 1000
3. 鋼筋的總數最多為 5000 根

1396ms

 1 class Solution {
 2     func tallestBillboard(_ rods: [Int]) -> Int {
 3         var n:Int = rods.count
 4         var h:Int = n/2
 5         var o:Int = 10002
 6         var ls:[Int] = [Int](repeating:-99999999,count:20005)
 7         for i in 0..<Int(pow(3, Double(h)))
 8         {
 9             var s:Int = 0
10             var ass:Int = 0
11             var v:Int = i
12             for j in 0..<h
13             {
14                 var w:Int = v % 3
15                 if w == 1
16                 {
17                     
18                 }
19                 else if w == 0
20                 {
21                     s += rods[j]
22                     ass += rods[j]
23                 }
24                 else
25                 {
26                     s -= rods[j]
27                     ass += rods[j]
28                 }
29                 v /= 3
30             }
31             ls[s+o] = max(ls[s+o], ass)
32         }
33         var ret:Int = 0
34         for i in 0..<Int(pow(3, Double(n - h)))
35         {
36             var s:Int = 0
37             var ass:Int = 0
38             var v:Int = i
39             for j in 0..<(n - h)
40             {
41                 var w:Int = v % 3
42                 if w == 1
43                 {
44                     
45                 }
46                 else if w == 0
47                 {
48                     s += rods[j + h]
49                     ass += rods[j + h]
50                 }
51                 else
52                 {
53                     s -= rods[j + h]
54                     ass += rods[j + h]
55                 }
56                 v /= 3
57             }
58             ret = max(ret, (ls[o-s] + ass) / 2)
59         }
60         return ret
61     }
62 }