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108 lines
4.4 KiB
Markdown
108 lines
4.4 KiB
Markdown
# [851. Loud and Rich](https://leetcode.com/problems/loud-and-rich/)
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## 题目
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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.
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For convenience, we'll call the person with label `x`, simply "person `x`".
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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.
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Also, we'll say `quiet = q` if person x has quietness `q`.
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Now, return `answer`, where `answer = 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`.
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**Example 1**:
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```
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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]
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Output: [5,5,2,5,4,5,6,7]
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Explanation:
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answer[0] = 5.
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Person 5 has more money than 3, which has more money than 1, which has more money than 0.
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The only person who is quieter (has lower quiet[x]) is person 7, but
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it isn't clear if they have more money than person 0.
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answer[7] = 7.
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Among all people that definitely have equal to or more money than person 7
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(which could be persons 3, 4, 5, 6, or 7), the person who is the quietest (has lower quiet[x])
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is person 7.
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The other answers can be filled out with similar reasoning.
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```
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**Note**:
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1. `1 <= quiet.length = N <= 500`
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2. `0 <= quiet[i] < N`, all `quiet[i]` are different.
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3. `0 <= richer.length <= N * (N-1) / 2`
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4. `0 <= richer[i][j] < N`
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5. `richer[i][0] != richer[i][1]`
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6. `richer[i]`'s are all different.
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7. The observations in `richer` are all logically consistent.
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## 题目大意
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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）。
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提示：
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- 1 <= quiet.length = N <= 500
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- 0 <= quiet[i] < N，所有 quiet[i] 都不相同。
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- 0 <= richer.length <= N * (N-1) / 2
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- 0 <= richer[i][j] < N
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- richer[i][0] != richer[i][1]
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- richer[i] 都是不同的。
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- 对 richer 的观察在逻辑上是一致的。
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## 解题思路
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- 给出 2 个数组，richer 和 quiet，要求输出 answer，其中 answer = y 的前提是，在所有拥有的钱不少于 x 的人中，y 是最安静的人（也就是安静值 quiet[y] 最小的人）
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- 由题意可知，`richer` 构成了一个有向无环图，首先使用字典建立图的关系，找到比当前下标编号富有的所有的人。然后使用广度优先层次遍历，不断的使用富有的人，但是安静值更小的人更新子节点即可。
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- 这一题还可以用拓扑排序来解答。将 `richer` 中描述的关系看做边，如果 `x > y`，则 `x` 指向 `y`。将 `quiet` 看成权值。用一个数组记录答案，初始时 `ans[i] = i`。然后对原图做拓扑排序，对于每一条边，如果发现 `quiet[ans[v]] > quiet[ans[u]]`，则 `ans[v]` 的答案为 `ans[u]`。时间复杂度即为拓扑排序的时间复杂度为 `O(m+n)`。空间复杂度需要 `O(m)` 的数组建图，需要 `O(n)` 的数组记录入度以及存储队列，所以空间复杂度为 `O(m+n)`。
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## 代码
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```go
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func loudAndRich(richer [][]int, quiet []int) []int {
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edges := make([][]int, len(quiet))
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for i := range edges {
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edges[i] = []int{}
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}
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indegrees := make([]int, len(quiet))
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for _, edge := range richer {
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n1, n2 := edge[0], edge[1]
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edges[n1] = append(edges[n1], n2)
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indegrees[n2]++
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}
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res := make([]int, len(quiet))
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for i := range res {
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res[i] = i
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}
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queue := []int{}
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for i, v := range indegrees {
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if v == 0 {
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queue = append(queue, i)
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}
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}
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for len(queue) > 0 {
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nexts := []int{}
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for _, n1 := range queue {
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for _, n2 := range edges[n1] {
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indegrees[n2]--
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if quiet[res[n2]] > quiet[res[n1]] {
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res[n2] = res[n1]
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}
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if indegrees[n2] == 0 {
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nexts = append(nexts, n2)
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}
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}
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}
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queue = nexts
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}
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return res
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}
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``` |