feat: add solutions to lcof2 problem: No.116,118

Author: yanglbme

lcof2/剑指 Offer II 116. 朋友圈/README.md

@@ -52,22 +52,254 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：深度优先搜索**
+
+判断城市之间是否属于同一个连通分量，最后连通分量的总数即为结果。
+
+**方法二：并查集**
+
+模板 1——朴素并查集：
+
+```python
+# 初始化，p存储每个点的祖宗节点
+p = [i for i in range(n)]
+# 返回x的祖宗节点
+def find(x):
+    if p[x] != x:
+        # 路径压缩
+        p[x] = find(p[x])
+    return p[x]
+# 合并a和b所在的两个集合
+p[find(a)] = find(b)
+```
+
+模板 2——维护 size 的并查集：
+
+```python
+# 初始化，p存储每个点的祖宗节点，size只有当节点是祖宗节点时才有意义，表示祖宗节点所在集合中，点的数量
+p = [i for i in range(n)]
+size = [1] * n
+# 返回x的祖宗节点
+def find(x):
+    if p[x] != x:
+        # 路径压缩
+        p[x] = find(p[x])
+    return p[x]
+# 合并a和b所在的两个集合
+size[find(b)] += size[find(a)]
+p[find(a)] = find(b)
+```
+
+模板 3——维护到祖宗节点距离的并查集：
+
+```python
+# 初始化，p存储每个点的祖宗节点，d[x]存储x到p[x]的距离
+p = [i for i in range(n)]
+d = [0] * n
+# 返回x的祖宗节点
+def find(x):
+    if p[x] != x:
+        t = find(p[x])
+        d[x] += d[p[x]]
+        p[x] = t
+    return p[x]
+# 合并a和b所在的两个集合
+p[find(a)] = find(b)
+d[find(a)] = dinstance
+```
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
+深度优先搜索：
+
 ```python
+class Solution:
+    def findCircleNum(self, isConnected: List[List[int]]) -> int:
+        def dfs(i):
+            for j in range(n):
+                if not visited[j] and isConnected[i][j] == 1:
+                    visited[j] = True
+                    dfs(j)
 
+        n = len(isConnected)
+        visited = [False] * n
+        num = 0
+        for i in range(n):
+            if not visited[i]:
+                dfs(i)
+                num += 1
+        return num
+```
+
+并查集：
+
+```python
+class Solution:
+    def findCircleNum(self, isConnected: List[List[int]]) -> int:
+        n = len(isConnected)
+        p = [i for i in range(n)]
+        
+        def find(x):
+            if p[x] != x:
+                p[x] = find(p[x])
+            return p[x]
+        
+        for i in range(n):
+            for j in range(n):
+                if i != j and isConnected[i][j] == 1:
+                    p[find(i)] = find(j)
+        return sum(i == find(i) for i in range(n))
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
+深度优先搜索：
+
 ```java
+class Solution {
+    public int findCircleNum(int[][] isConnected) {
+        int n = isConnected.length;
+        boolean[] visited = new boolean[n];
+        int num = 0;
+        for (int i = 0; i < n; ++i) {
+            if (!visited[i]) {
+                dfs(isConnected, visited, i, n);
+                ++num;
+            }
+        }
+        return num;
+    }
+
+    private void dfs(int[][] isConnected, boolean[] visited, int i, int n) {
+        for (int j = 0; j < n; ++j) {
+            if (!visited[j] && isConnected[i][j] == 1) {
+                visited[j] = true;
+                dfs(isConnected, visited, j, n);
+            }
+        }
+    }
+}
+```
+
+并查集：
+
+```java
+class Solution {
+    private int[] p;
+
+    public int findCircleNum(int[][] isConnected) {
+        int n = isConnected.length;
+        p = new int[n];
+        for (int i = 0; i < n; ++i) {
+            p[i] = i;
+        }
+        for (int i = 0; i < n; ++i) {
+            for (int j = 0; j < n; ++j) {
+                if (isConnected[i][j] == 1) {
+                    p[find(i)] = find(j);
+                }
+            }
+        }
+        int cnt = 0;
+        for (int i = 0; i < n; ++i) {
+            if (i == find(i)) {
+                ++cnt;
+            }
+        }
+        return cnt;
+    }
+
+    private int find(int x) {
+        if (p[x] != x) {
+            p[x] = find(p[x]);
+        }
+        return p[x];
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    vector<int> p;
+
+    int findCircleNum(vector<vector<int>> &isConnected) {
+        int n = isConnected.size();
+        p.resize(n);
+        for (int i = 0; i < n; ++i)
+        {
+            p[i] = i;
+        }
+        for (int i = 0; i < n; ++i)
+        {
+            for (int j = 0; j < n; ++j)
+            {
+                if (isConnected[i][j])
+                {
+                    p[find(i)] = find(j);
+                }
+            }
+        }
+        int cnt = 0;
+        for (int i = 0; i < n; ++i)
+        {
+            if (i == find(i))
+                ++cnt;
+        }
+        return cnt;
+    }
+
+    int find(int x) {
+        if (p[x] != x)
+        {
+            p[x] = find(p[x]);
+        }
+        return p[x];
+    }
+};
+```
+
+### **Go**
+
+```go
+var p []int
+
+func findCircleNum(isConnected [][]int) int {
+	n := len(isConnected)
+	p = make([]int, n)
+	for i := 1; i < n; i++ {
+		p[i] = i
+	}
+	for i := 0; i < n; i++ {
+		for j := 0; j < n; j++ {
+			if isConnected[i][j] == 1 {
+				p[find(i)] = find(j)
+			}
+		}
+	}
+	cnt := 0
+	for i := 0; i < n; i++ {
+		if i == find(i) {
+			cnt++
+		}
+	}
+	return cnt
+}
 
+func find(x int) int {
+	if p[x] != x {
+		p[x] = find(p[x])
+	}
+	return p[x]
+}
 ```
 
 ### **...**

lcof2/剑指 Offer II 116. 朋友圈/Solution.cpp

@@ -0,0 +1,38 @@
+class Solution {
+public:
+    vector<int> p;
+
+    int findCircleNum(vector<vector<int>> &isConnected) {
+        int n = isConnected.size();
+        p.resize(n);
+        for (int i = 0; i < n; ++i)
+        {
+            p[i] = i;
+        }
+        for (int i = 0; i < n; ++i)
+        {
+            for (int j = 0; j < n; ++j)
+            {
+                if (isConnected[i][j])
+                {
+                    p[find(i)] = find(j);
+                }
+            }
+        }
+        int cnt = 0;
+        for (int i = 0; i < n; ++i)
+        {
+            if (i == find(i))
+                ++cnt;
+        }
+        return cnt;
+    }
+
+    int find(int x) {
+        if (p[x] != x)
+        {
+            p[x] = find(p[x]);
+        }
+        return p[x];
+    }
+};

lcof2/剑指 Offer II 116. 朋友圈/Solution.go

@@ -0,0 +1,30 @@
+var p []int
+
+func findCircleNum(isConnected [][]int) int {
+	n := len(isConnected)
+	p = make([]int, n)
+	for i := 1; i < n; i++ {
+		p[i] = i
+	}
+	for i := 0; i < n; i++ {
+		for j := 0; j < n; j++ {
+			if isConnected[i][j] == 1 {
+				p[find(i)] = find(j)
+			}
+		}
+	}
+	cnt := 0
+	for i := 0; i < n; i++ {
+		if i == find(i) {
+			cnt++
+		}
+	}
+	return cnt
+}
+
+func find(x int) int {
+	if p[x] != x {
+		p[x] = find(p[x])
+	}
+	return p[x]
+}

lcof2/剑指 Offer II 116. 朋友圈/Solution.java

@@ -0,0 +1,23 @@
+class Solution {
+    public int findCircleNum(int[][] isConnected) {
+        int n = isConnected.length;
+        boolean[] visited = new boolean[n];
+        int num = 0;
+        for (int i = 0; i < n; ++i) {
+            if (!visited[i]) {
+                dfs(isConnected, visited, i, n);
+                ++num;
+            }
+        }
+        return num;
+    }
+
+    private void dfs(int[][] isConnected, boolean[] visited, int i, int n) {
+        for (int j = 0; j < n; ++j) {
+            if (!visited[j] && isConnected[i][j] == 1) {
+                visited[j] = true;
+                dfs(isConnected, visited, j, n);
+            }
+        }
+    }
+}

lcof2/剑指 Offer II 116. 朋友圈/Solution.py

@@ -0,0 +1,16 @@
+class Solution:
+    def findCircleNum(self, isConnected: List[List[int]]) -> int:
+        def dfs(i):
+            for j in range(n):
+                if not visited[j] and isConnected[i][j] == 1:
+                    visited[j] = True
+                    dfs(j)
+        
+        n = len(isConnected)
+        visited = [False] * n
+        num = 0
+        for i in range(n):
+            if not visited[i]:
+                dfs(i)
+                num += 1
+        return num