Micro fix

leetcode/0547.Number-of-Provinces/547. Number of Provinces.go

@@ -0,0 +1,49 @@
+package leetcode
+
+import (
+	"github.com/halfrost/LeetCode-Go/template"
+)
+
+// 解法一 并查集
+
+func findCircleNum(M [][]int) int {
+	n := len(M)
+	if n == 0 {
+		return 0
+	}
+	uf := template.UnionFind{}
+	uf.Init(n)
+	for i := 0; i < n; i++ {
+		for j := 0; j <= i; j++ {
+			if M[i][j] == 1 {
+				uf.Union(i, j)
+			}
+		}
+	}
+	return uf.TotalCount()
+}
+
+// 解法二 FloodFill DFS 暴力解法
+func findCircleNum1(M [][]int) int {
+	if len(M) == 0 {
+		return 0
+	}
+	visited := make([]bool, len(M))
+	res := 0
+	for i := range M {
+		if !visited[i] {
+			dfs547(M, i, visited)
+			res++
+		}
+	}
+	return res
+}
+
+func dfs547(M [][]int, cur int, visited []bool) {
+	visited[cur] = true
+	for j := 0; j < len(M[cur]); j++ {
+		if !visited[j] && M[cur][j] == 1 {
+			dfs547(M, j, visited)
+		}
+	}
+}

leetcode/0547.Number-of-Provinces/547. Number of Provinces_test.go

@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question547 struct {
+	para547
+	ans547
+}
+
+// para 是参数
+// one 代表第一个参数
+type para547 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans547 struct {
+	one int
+}
+
+func Test_Problem547(t *testing.T) {
+
+	qs := []question547{
+
+		{
+			para547{[][]int{{0, 0, 0}, {0, 1, 0}, {0, 0, 0}}},
+			ans547{3},
+		},
+
+		{
+			para547{[][]int{{1, 1, 0}, {1, 1, 0}, {0, 0, 1}}},
+			ans547{2},
+		},
+
+		{
+			para547{[][]int{{1, 1, 0}, {1, 1, 1}, {0, 1, 1}}},
+			ans547{1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 547------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans547, q.para547
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findCircleNum(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0547.Number-of-Provinces/README.md

@@ -0,0 +1,54 @@
+# [547. Number of Provinces](https://leetcode.com/problems/number-of-provinces/)
+
+## 题目
+
+There are **N** students in a class. Some of them are friends, while some are not. Their friendship is transitive in nature. For example, if A is a **direct** friend of B, and B is a **direct**friend of C, then A is an **indirect** friend of C. And we defined a friend circle is a group of students who are direct or indirect friends.
+
+Given a **N*N** matrix **M** representing the friend relationship between students in the class. If M[i][j] = 1, then the ith and jth students are **direct** friends with each other, otherwise not. And you have to output the total number of friend circles among all the students.
+
+**Example 1:**
+
+    Input: 
+    [[1,1,0],
+     [1,1,0],
+     [0,0,1]]
+    Output: 2
+    Explanation:The 0th and 1st students are direct friends, so they are in a friend circle. 
+    The 2nd student himself is in a friend circle. So return 2.
+
+**Example 2:**
+
+    Input: 
+    [[1,1,0],
+     [1,1,1],
+     [0,1,1]]
+    Output: 1
+    Explanation:The 0th and 1st students are direct friends, the 1st and 2nd students are direct friends, 
+    so the 0th and 2nd students are indirect friends. All of them are in the same friend circle, so return 1.
+
+**Note:**
+
+1. N is in range [1,200].
+2. M[i][i] = 1 for all students.
+3. If M[i][j] = 1, then M[j][i] = 1.
+
+
+## 题目大意
+
+班上有 N 名学生。其中有些人是朋友，有些则不是。他们的友谊具有是传递性。如果已知 A 是 B 的朋友，B 是 C 的朋友，那么我们可以认为 A 也是 C 的朋友。所谓的朋友圈，是指所有朋友的集合。
+
+给定一个 N * N 的矩阵 M，表示班级中学生之间的朋友关系。如果 M[i][j] = 1，表示已知第 i 个和 j 个学生互为朋友关系，否则为不知道。你必须输出所有学生中的已知的朋友圈总数。
+
+
+注意：
+
+- N 在[1,200]的范围内。
+- 对于所有学生，有M[i][i] = 1。
+- 如果有 M[i][j] = 1，则有 M[j][i] = 1。
+
+
+## 解题思路
+
+
+- 给出一个二维矩阵，矩阵中的行列表示的是两个人之间是否是朋友关系，如果是 1，代表两个人是朋友关系。由于自己和自肯定朋友关系，所以对角线上都是 1，并且矩阵也是关于从左往右下的这条对角线对称。
+- 这题有 2 种解法，第一种解法是并查集，依次扫描矩阵，如果两个人认识，并且 root 并不相等就执行 union 操作。扫完所有矩阵，最后数一下还有几个不同的 root 就是最终答案。第二种解法是 DFS 或者 BFS。利用 FloodFill 的想法去染色，每次染色一次，计数器加一。最终扫完整个矩阵，计数器的结果就是最终结果。