[Swift]LeetCode851.喧闹和富有|LoudandRich

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In a group of N people (labelled 0, 1, 2, ..., N-1), each person has different amounts of money, and different levels of quietness.

For convenience, we'll call the person with label x, simply "person x".

We'll say that richer[i] = [x, y] if person x definitely has more money than person y. Note that richer may only be a subset of valid observations.

Also, we'll say quiet[x] = q if person x has quietness q.

Now, return answer, where answer[x] = y if y is the least quiet person (that is, the person y with the smallest value of quiet[y]), among all people who definitely have equal to or more money than person x.

Example 1:

Input: richer = [3,2,5,4,6,1,7,0]
Output: [5,5,2,5,4,5,6,7]
Explanation: 
answer[0] = 5.
Person 5 has more money than 3, which has more money than 1, which has more money than 0.
The only person who is quieter (has lower quiet[x]) is person 7, but
it isn't clear if they have more money than person 0.

answer[7] = 7.
Among all people that definitely have equal to or more money than person 7
(which could be persons 3, 4, 5, 6, or 7), the person who is the quietest (has lower quiet[x])
is person 7.

The other answers can be filled out with similar reasoning.

Note:

1 <= quiet.length = N <= 500
0 <= quiet[i] < N, all quiet[i] are different.
0 <= richer.length <= N * (N-1) / 2
0 <= richer[i][j] < N
richer[i][0] != richer[i][1]
richer[i]'s are all different.
The observations in richer are all logically consistent.

在一组 N 个人（编号为 0, 1, 2, ..., N-1）中，每个人都有不同数目的钱，以及不同程度的安静（quietness）。

为了方便起见，我们将编号为 x 的人简称为 "person x "。

如果能够肯定 person x 比 person y 更有钱的话，我们会说 richer[i] = [x, y] 。注意 richer 可能只是有效观察的一个子集。

另外，如果 person x 的安静程度为 q ，我们会说 quiet[x] = q 。

现在，返回答案 answer ，其中 answer[x] = y 的前提是，在所有拥有的钱不少于 person x 的人中，person y 是最安静的人（也就是安静值 quiet[y] 最小的人）。

示例：

输入：richer = [[1,0],[2,1],[3,1],[3,7],[4,3],[5,3],[6,3]], quiet = [3,2,5,4,6,1,7,0]
输出：[5,5,2,5,4,5,6,7]
解释： 
answer[0] = 5，
person 5 比 person 3 有更多的钱，person 3 比 person 1 有更多的钱，person 1 比 person 0 有更多的钱。
唯一较为安静（有较低的安静值 quiet[x]）的人是 person 7，
但是目前还不清楚他是否比 person 0 更有钱。

answer[7] = 7，
在所有拥有的钱肯定不少于 person 7 的人中(这可能包括 person 3，4，5，6 以及 7)，
最安静(有较低安静值 quiet[x])的人是 person 7。

其他的答案也可以用类似的推理来解释。

提示：

1 <= quiet.length = N <= 500
0 <= quiet[i] < N，所有 quiet[i] 都不相同。
0 <= richer.length <= N * (N-1) / 2
0 <= richer[i][j] < N
richer[i][0] != richer[i][1]
richer[i] 都是不同的。
对 richer 的观察在逻辑上是一致的。

Runtime: 488 ms

Memory Usage: 21.4 MB

 1 class Solution {
 2     var richer2:[Int:[Int]] = [Int:[Int]]()
 3     var res:[Int] = [Int]()
 4     func loudAndRich(_ richer: [[Int]], _ quiet: [Int]) -> [Int] {
 5         var n:Int = quiet.count
 6         for i in 0..<n
 7         {
 8             richer2[i] = [Int]()            
 9         }
10         for v in richer
11         {
12             richer2[v[1],default:[Int]()].append(v[0])            
13         }
14         res = [Int](repeating:-1,count:n)
15         for i in 0..<n
16         {
17             dfs(i, quiet)
18         }
19         return res
20     }
21     
22     func dfs(_ i:Int,_ quiet:[Int]) -> Int
23     {
24         if res[i] >= 0 {return res[i]}
25         res[i] = i
26         for j in richer2[i,default:[Int]()]
27         {
28             if quiet[res[i]] > quiet[dfs(j, quiet)]
29             {
30                 res[i] = res[j]
31             }
32         }
33         return res[i]
34     }
35 }

Runtime: 540 ms

Memory Usage: 22.2 MB

 1 class Solution {
 2     func loudAndRich(_ richer: [[Int]], _ quiet: [Int]) -> [Int] {       
 3         //这是个列表，
 4         var mans:[Man] = (0 ..< quiet.count).map{Man.init(index: $0, quiet: quiet[$0],quiets: quiet)}
 5         //做成单向链表 叫这个名字吧？
 6         for paired in richer {            mans[paired.last!].richs.insert(mans[paired.first!])
 7         }        
 8         //存 结果答案
 9         var result:[Int] = []
10         //遍历答案
11         for e in mans {            
12             result.append(
13                 e.mostQuietMan.index
14             )            
15         }
16         return result        
17     }
18 }
19 
20 /// 抽象的人
21 class Man {
22     var index:Int //自己的位置
23     var quiet:Int //安静度    
24     var quiets:[Int]
25     var richs:Set<Man> = [] //比自己富的人，用set可去重，有层数
26     init(index:Int,quiet:Int, quiets: [Int]){
27         self.index = index
28         self.quiet = quiet
29         self.quiets = quiets
30     }
31     lazy var mostQuietMan = getmostQuietMan(quiet: self.quiets)
32     func getmostQuietMan(quiet: [Int]) -> Man {
33         var mostQuietMan = self
34         for man in self.richs{
35             if quiet[man.index] < quiet[mostQuietMan.index]{
36                 mostQuietMan = man
37             }
38             if quiet[man.mostQuietMan.index] < quiet[mostQuietMan.index] {
39                 mostQuietMan = man.mostQuietMan
40             }
41         }
42         return mostQuietMan
43     }    
44 }
45 
46 //Set需要遵循hashable
47 extension Man:Hashable {
48     static func == (lhs: Man, rhs: Man) -> Bool {
49         return lhs.index == rhs.index
50     }
51     
52     var hashValue:Int {
53         return self.index
54     }
55 }

Runtime: 3172 ms

Memory Usage: 34.3 MB

 1 class Solution {
 2     func loudAndRich(_ richer: [[Int]], _ quiet: [Int]) -> [Int] {
 3         var mans:[Man] = (0 ..< quiet.count).map{  Man.init(index: $0, quiet: quiet[$0])}
 4         //做成单向链表 叫这个名字吧？
 5         for paired in richer {
 6             mans[paired.last!].richs.insert(mans[paired.first!])
 7         }
 8         var result:[Int] = []
 9         
10         for e in mans {            
11             result.append(
12             e.richers.min { (a, b) -> Bool in
13                return a.quiet < b.quiet
14             }!.index
15             )            
16         }               
17         return result        
18     }
19 }
20 
21 class Man {
22     var index:Int
23     var quiet:Int
24     var richs:Set<Man> = []
25     init(index:Int,quiet:Int){
26         self.index = index
27         self.quiet = quiet
28     }
29     lazy var richers:Set<Man> = self.getRichers()
30     func getRichers() -> Set<Man> {        
31             var rs:Set<Man> = richs
32             for e in richs {
33                 rs = rs.union(e.richers)
34             }
35             rs.insert(self)
36             return rs        
37     }    
38 }
39 
40 extension Man:Hashable {
41     static func == (lhs: Man, rhs: Man) -> Bool {
42         return lhs.index == rhs.index
43     }
44     
45     var hashValue:Int {
46         return self.index
47     }
48 }