规范格式

leetcode/0851.Loud-and-Rich/851. Loud and Rich.go

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+package leetcode
+
+func loudAndRich(richer [][]int, quiet []int) []int {
+	edges := make([][]int, len(quiet))
+	for i := range edges {
+		edges[i] = []int{}
+	}
+	indegrees := make([]int, len(quiet))
+	for _, edge := range richer {
+		n1, n2 := edge[0], edge[1]
+		edges[n1] = append(edges[n1], n2)
+		indegrees[n2]++
+	}
+	res := make([]int, len(quiet))
+	for i := range res {
+		res[i] = i
+	}
+	queue := []int{}
+	for i, v := range indegrees {
+		if v == 0 {
+			queue = append(queue, i)
+		}
+	}
+	for len(queue) > 0 {
+		nexts := []int{}
+		for _, n1 := range queue {
+			for _, n2 := range edges[n1] {
+				indegrees[n2]--
+				if quiet[res[n2]] > quiet[res[n1]] {
+					res[n2] = res[n1]
+				}
+				if indegrees[n2] == 0 {
+					nexts = append(nexts, n2)
+				}
+			}
+		}
+		queue = nexts
+	}
+	return res
+}

leetcode/0851.Loud-and-Rich/851. Loud and Rich_test.go

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+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question851 struct {
+	para851
+	ans851
+}
+
+// para 是参数
+// one 代表第一个参数
+type para851 struct {
+	richer [][]int
+	quiet  []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans851 struct {
+	one []int
+}
+
+func Test_Problem851(t *testing.T) {
+
+	qs := []question851{
+
+		question851{
+			para851{[][]int{[]int{1, 0}, []int{2, 1}, []int{3, 1}, []int{3, 7}, []int{4, 3}, []int{5, 3}}, []int{3, 2, 5, 4, 6, 1, 7, 0}},
+			ans851{[]int{5, 5, 2, 5, 4, 5, 6, 7}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 851------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans851, q.para851
+		fmt.Printf("【input】:%v       【output】:%v\n", p, loudAndRich(p.richer, p.quiet))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0851.Loud-and-Rich/README.md

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+# 851. Loud and Rich
+
+
+## 题目
+
+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 = q` if person x has quietness `q`.
+
+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`.
+
+**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.
+
+## 题目大意
+
+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）。
+
+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。
+
+
+## 解题思路
+
+- 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）
+- 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。
+- 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)`。
+
+## 代码
+
+```go
+func loudAndRich(richer [][]int, quiet []int) []int {
+	edges := make([][]int, len(quiet))
+	for i := range edges {
+		edges[i] = []int{}
+	}
+	indegrees := make([]int, len(quiet))
+	for _, edge := range richer {
+		n1, n2 := edge[0], edge[1]
+		edges[n1] = append(edges[n1], n2)
+		indegrees[n2]++
+	}
+	res := make([]int, len(quiet))
+	for i := range res {
+		res[i] = i
+	}
+	queue := []int{}
+	for i, v := range indegrees {
+		if v == 0 {
+			queue = append(queue, i)
+		}
+	}
+	for len(queue) > 0 {
+		nexts := []int{}
+		for _, n1 := range queue {
+			for _, n2 := range edges[n1] {
+				indegrees[n2]--
+				if quiet[res[n2]] > quiet[res[n1]] {
+					res[n2] = res[n1]
+				}
+				if indegrees[n2] == 0 {
+					nexts = append(nexts, n2)
+				}
+			}
+		}
+		queue = nexts
+	}
+	return res
+}
+```