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.
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]
answer = 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.
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.
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] != richer[i]
richer[i]’s are all different.
The observations in richer are all logically consistent.
可以看到2，4，5，6入度为0，即没有比他们更多钱的人，那么对于这种节点 ,ans[i] = i
对于3,则要从指向他的节点（4，5，6）以及他自身中找到quiet值最小的点index,ans = min(3,4,5,6,key=lambda x: quiet[ans[x]]) ,依次类推即可