Merge pull request #216 from gostool/leetcode0997

Leetcode0997

leetcode/0997.Find-the-Town-Judge/997.Find the Town Judge.go

@@ -0,0 +1,22 @@
+package leetcode
+
+func findJudge(n int, trust [][]int) int {
+	if n == 1 && len(trust) == 0 {
+		return 1
+	}
+	judges := make(map[int]int)
+	for _, v := range trust {
+		judges[v[1]] += 1
+	}
+	for _, v := range trust {
+		if _, ok := judges[v[0]]; ok {
+			delete(judges, v[0])
+		}
+	}
+	for k, v := range judges {
+		if v == n-1 {
+			return k
+		}
+	}
+	return -1
+}

leetcode/0997.Find-the-Town-Judge/997.Find the Town Judge_test.go

@@ -0,0 +1,51 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question997 struct {
+	para997
+	ans997
+}
+
+// para 是参数
+type para997 struct {
+	n     int
+	trust [][]int
+}
+
+// ans 是答案
+type ans997 struct {
+	ans int
+}
+
+func Test_Problem997(t *testing.T) {
+
+	qs := []question997{
+
+		{
+			para997{2, [][]int{{1, 2}}},
+			ans997{2},
+		},
+
+		{
+			para997{3, [][]int{{1, 3}, {2, 3}}},
+			ans997{3},
+		},
+
+		{
+			para997{3, [][]int{{1, 3}, {2, 3}, {3, 1}}},
+			ans997{-1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 997------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans997, q.para997
+		fmt.Printf("【input】:%v    【output】:%v\n", p, findJudge(p.n, p.trust))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0997.Find-the-Town-Judge/README.md

@@ -0,0 +1,87 @@
+# [997. Find the Town Judge](https://leetcode-cn.com/problems/find-the-town-judge/)
+
+## 题目
+
+In a town, there are n people labeled from 1 to n. There is a rumor that one of these people is secretly the town judge.
+
+If the town judge exists, then:
+
+- The town judge trusts nobody.
+- Everybody (except for the town judge) trusts the town judge.
+- There is exactly one person that satisfies properties 1 and 2.
+
+You are given an array trust where trust[i] = [ai, bi] representing that the person labeled ai trusts the person labeled bi.
+
+Return the label of the town judge if the town judge exists and can be identified, or return -1 otherwise.
+
+**Example 1**:
+
+    Input: n = 2, trust = [[1,2]]
+    Output: 2
+
+**Example 2**:
+
+    Input: n = 3, trust = [[1,3],[2,3]]
+    Output: 3
+
+**Example 3**:
+
+    Input: n = 3, trust = [[1,3],[2,3],[3,1]]
+    Output: -1
+
+**Constraints:**
+
+- 1 <= n <= 1000
+- 0 <= trust.length <= 10000
+- trust[i].length == 2
+- All the pairs of trust are unique.
+- ai != bi
+- 1 <= ai, bi <= n
+
+## 题目大意
+
+小镇里有 n 个人，按从 1 到 n 的顺序编号。传言称，这些人中有一个暗地里是小镇法官。
+
+如果小镇法官真的存在，那么：
+
+- 小镇法官不会信任任何人。
+- 每个人（除了小镇法官）都信任这位小镇法官。
+- 只有一个人同时满足属性 1 和属性 2 。
+
+给你一个数组 trust ，其中 trust[i] = [ai, bi] 表示编号为 ai 的人信任编号为 bi 的人。
+
+如果小镇法官存在并且可以确定他的身份，请返回该法官的编号；否则，返回 -1 。
+
+## 解题思路
+
+入度和出度统计
+
+- 被人信任定义为入度, 信任别人定义为出度
+- 如果1-n之间有数字x的入度为n - 1，出度为0，则返回x
+
+## 代码
+
+```go
+package leetcode
+
+func findJudge(n int, trust [][]int) int {
+	if n == 1 && len(trust) == 0 {
+		return 1
+	}
+	judges := make(map[int]int)
+	for _, v := range trust {
+		judges[v[1]] += 1
+	}
+	for _, v := range trust {
+		if _, ok := judges[v[0]]; ok {
+			delete(judges, v[0])
+		}
+	}
+	for k, v := range judges {
+		if v == n-1 {
+			return k
+		}
+	}
+	return -1
+}
+```