[Swift]LeetCode851. 喧闹和富有 | Loud and Rich

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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 = [[1,0],[2,1],[3,1],[3,7],[4,3],[5,3],[6,3]], quiet = [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. 1 <= quiet.length = N <= 500
2. 0 <= quiet[i] < N, all quiet[i] are different.
3. 0 <= richer.length <= N * (N-1) / 2
4. 0 <= richer[i][j] < N
5. richer[i][0] != richer[i][1]
6. richer[i]'s are all different.
7. 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. 1 <= quiet.length = N <= 500
2. 0 <= quiet[i] < N，所有 quiet[i] 都不相同。
3. 0 <= richer.length <= N * (N-1) / 2
4. 0 <= richer[i][j] < N
5. richer[i][0] != richer[i][1]
6. richer[i] 都是不同的。
7. 对 richer 的观察在逻辑上是一致的。 

---

Runtime: 488 ms

Memory Usage: 21.4 MB

class Solution {
    var richer2:[Int:[Int]] = [Int:[Int]]()
    var res:[Int] = [Int]()
    func loudAndRich(_ richer: [[Int]], _ quiet: [Int]) -> [Int] {
        var n:Int = quiet.count
        for i in 0..<n
        {
            richer2[i] = [Int]()            
        }
        for v in richer
        {
            richer2[v[1],default:[Int]()].append(v[0])            
        }
        res = [Int](repeating:-1,count:n)
        for i in 0..<n
        {
            dfs(i, quiet)
        }
        return res
    }
    
    func dfs(_ i:Int,_ quiet:[Int]) -> Int
    {
        if res[i] >= 0 {return res[i]}
        res[i] = i
        for j in richer2[i,default:[Int]()]
        {
            if quiet[res[i]] > quiet[dfs(j, quiet)]
            {
                res[i] = res[j]
            }
        }
        return res[i]
    }
}

---

Runtime: 540 ms

Memory Usage: 22.2 MB

class Solution {
    func loudAndRich(_ richer: [[Int]], _ quiet: [Int]) -> [Int] {       
        //这是个列表，
        var mans:[Man] = (0 ..< quiet.count).map{Man.init(index: $0, quiet: quiet[$0],quiets: quiet)}
        //做成单向链表 叫这个名字吧？
        for paired in richer {            mans[paired.last!].richs.insert(mans[paired.first!])
        }        
        //存 结果答案
        var result:[Int] = []
        //遍历答案
        for e in mans {            
            result.append(
                e.mostQuietMan.index
            )            
        }
        return result        
    }
}

/// 抽象的人
class Man {
    var index:Int //自己的位置
    var quiet:Int //安静度    
    var quiets:[Int]
    var richs:Set<Man> = [] //比自己富的人，用set可去重，有层数
    init(index:Int,quiet:Int, quiets: [Int]){
        self.index = index
        self.quiet = quiet
        self.quiets = quiets
    }
    lazy var mostQuietMan = getmostQuietMan(quiet: self.quiets)
    func getmostQuietMan(quiet: [Int]) -> Man {
        var mostQuietMan = self
        for man in self.richs{
            if quiet[man.index] < quiet[mostQuietMan.index]{
                mostQuietMan = man
            }
            if quiet[man.mostQuietMan.index] < quiet[mostQuietMan.index] {
                mostQuietMan = man.mostQuietMan
            }
        }
        return mostQuietMan
    }    
}

//Set需要遵循hashable
extension Man:Hashable {
    static func == (lhs: Man, rhs: Man) -> Bool {
        return lhs.index == rhs.index
    }
    
    var hashValue:Int {
        return self.index
    }
}

---

Runtime: 3172 ms

Memory Usage: 34.3 MB

class Solution {
    func loudAndRich(_ richer: [[Int]], _ quiet: [Int]) -> [Int] {
        var mans:[Man] = (0 ..< quiet.count).map{  Man.init(index: $0, quiet: quiet[$0])}
        //做成单向链表 叫这个名字吧？
        for paired in richer {
            mans[paired.last!].richs.insert(mans[paired.first!])
        }
        var result:[Int] = []
        
        for e in mans {            
            result.append(
            e.richers.min { (a, b) -> Bool in
               return a.quiet < b.quiet
            }!.index
            )            
        }               
        return result        
    }
}

class Man {
    var index:Int
    var quiet:Int
    var richs:Set<Man> = []
    init(index:Int,quiet:Int){
        self.index = index
        self.quiet = quiet
    }
    lazy var richers:Set<Man> = self.getRichers()
    func getRichers() -> Set<Man> {        
            var rs:Set<Man> = richs
            for e in richs {
                rs = rs.union(e.richers)
            }
            rs.insert(self)
            return rs        
    }    
}

extension Man:Hashable {
    static func == (lhs: Man, rhs: Man) -> Bool {
        return lhs.index == rhs.index
    }
    
    var hashValue:Int {
        return self.index
    }
}