Merge pull request #69 from saenaii/develop

add leetcode547

src/0547.Friend-Circles/Solution.go

@@ -0,0 +1,49 @@
+package Solution
+
+var p []int
+var r []int
+var ans int
+
+func findCircleNum(M [][]int) int {
+    ans = len(M)
+    initialize(ans)
+
+    for i := 0; i < len(M); i++ {
+        for j := 0; j < i; j++ {
+            if 1 == M[i][j] {
+                union(i, j)
+            }
+        }
+    }
+    return ans
+}
+
+func initialize(l int) {
+    p = make([]int, l)
+    r = make([]int, l)
+    for i, _ := range p {
+        p[i] = i
+        r[i] = 1
+    }
+}
+
+func find(x int) int {
+    if x != p[x] {
+        p[x] = find(p[x])
+    }
+    return p[x]
+}
+
+func union(x, y int) {
+    x, y = find(x), find(y)
+    if x != y {
+        if r[x] <= r[y] {
+            p[x] = y
+            r[y] += r[x]
+        } else {
+            p[y] = x
+            r[x] += r[y]
+        }
+        ans--
+    }
+}

src/0547.Friend-Circles/Solution_test.go

@@ -0,0 +1,44 @@
+package Solution
+
+import (
+	"reflect"
+	"testing"
+)
+
+func TestSolution(t *testing.T) {
+	// 测试用例
+	cases := []struct {
+		name 	string
+		inputs	[][]int
+		expect 	int
+	}{
+		{"TestCases 1",
+			[][]int{
+				{1,1,0},
+				{1,1,1},
+				{0,1,1},
+			},
+			1,
+		},
+	}
+
+	// 开始测试
+	for _, c := range cases {
+		t.Run(c.name, func(t *testing.T) {
+			got := findCircleNum(c.inputs)
+			if !reflect.DeepEqual(got, c.expect) {
+				t.Fatalf("expected: %v, but got: %v, with inputs: %v", c.expect, got, c.inputs)
+			}
+		})
+	}
+}
+
+//	压力测试
+func BenchmarkSolution(b *testing.B) {
+
+}
+
+//	使用案列
+func ExampleSolution() {
+
+}

src/0547.Friend-Circles/readme.md

@@ -0,0 +1,95 @@
+# [547. Friend Circles][title]
+
+## Description
+
+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.
+```
+
+## 题意
+> 判断几个学生是否在同一个圈子。
+
+## 题解
+
+### 思路 1
+> 使用并查集
+
+```go
+var p []int
+var r []int
+var ans int
+
+func findCircleNum(M [][]int) int {
+    ans = len(M)
+    initialize(ans)
+
+    for i := 0; i < len(M); i++ {
+        for j := 0; j < i; j++ {
+            if 1 == M[i][j] {
+                union(i, j)
+            }
+        }
+    }
+    return ans
+}
+
+func initialize(l int) {
+    p = make([]int, l)
+    r = make([]int, l)
+    for i, _ := range p {
+        p[i] = i
+        r[i] = 1
+    }
+    fmt.Println(p, r)
+}
+
+func find(x int) int {
+    if x != p[x] {
+        p[x] = find(p[x])
+    }
+    return p[x]
+}
+
+func union(x, y int) {
+    x, y = find(x), find(y)
+    if x != y {
+        if r[x] <= r[y] {
+            p[x] = y
+            r[y] += r[x]
+        } else {
+            p[y] = x
+            r[x] += r[y]
+        }
+        ans--
+    }
+}
+```
+
+## 结语
+
+如果你同我一样热爱数据结构、算法、LeetCode，可以关注我 GitHub 上的 LeetCode 题解：[awesome-golang-leetcode][me]
+
+[title]: https://leetcode.com/problems/friend-circles/description/
+[me]: https://github.com/kylesliu/awesome-golang-leetcode