feat: add solutions to lc/lcof2 problem: Is Graph Bipartite

Author: yanglbme

lcof2/剑指 Offer II 106. 二分图/README.md

@@ -62,22 +62,187 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+并查集。
+
+并查集模板：
+
+模板 1——朴素并查集：
+
+```python
+# 初始化，p存储每个点的父节点
+p = list(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存储每个点的父节点，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 = list(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
+if find(a) != find(b):
+    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存储每个点的父节点，d[x]存储x到p[x]的距离
+p = list(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)] = distance
+```
+
+对于本题，如果是二分图，那么图中每个顶点的所有邻接点都应该属于同一集合，且不与顶点处于同一集合，因此我们可以使用并查集。遍历图中每个顶点，如果发现存在当前顶点与对应的邻接点处于同一个集合，说明不是二分图。否则将当前节点的邻接点相互进行合并。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def isBipartite(self, graph: List[List[int]]) -> bool:
+        n = len(graph)
+        p = list(range(n))
+
+        def find(x):
+            if p[x] != x:
+                p[x] = find(p[x])
+            return p[x]
+
+        for u, g in enumerate(graph):
+            for v in g:
+                if find(u) == find(v):
+                    return False
+                p[find(v)] = find(g[0])
+        return True
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    private int[] p;
+
+    public boolean isBipartite(int[][] graph) {
+        int n = graph.length;
+        p = new int[n];
+        for (int i = 0; i < n; ++i) {
+            p[i] = i;
+        }
+        for (int u = 0; u < n; ++u) {
+            int[] g = graph[u];
+            for (int v : g) {
+                if (find(u) == find(v)) {
+                    return false;
+                }
+                p[find(v)] = find(g[0]);
+            }
+        }
+        return true;
+    }
+
+    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;
+
+    bool isBipartite(vector<vector<int>>& graph) {
+        int n = graph.size();
+        p.resize(n);
+        for (int i = 0; i < n; ++i) p[i] = i;
+        for (int u = 0; u < n; ++u)
+        {
+            auto g = graph[u];
+            for (int v : g)
+            {
+                if (find(u) == find(v)) return false;
+                p[find(v)] = find(g[0]);
+            }
+        }
+        return true;
+    }
+
+    int find(int x) {
+        if (p[x] != x) p[x] = find(p[x]);
+        return p[x];
+    }
+};
+```
 
+### **Go**
+
+```go
+var p []int
+
+func isBipartite(graph [][]int) bool {
+	n := len(graph)
+	p = make([]int, n)
+	for i := 0; i < n; i++ {
+		p[i] = i
+	}
+	for u, g := range graph {
+		for _, v := range g {
+			if find(u) == find(v) {
+				return false
+			}
+			p[find(v)] = find(g[0])
+		}
+	}
+	return true
+}
+
+func find(x int) int {
+	if p[x] != x {
+		p[x] = find(p[x])
+	}
+	return p[x]
+}
 ```
 
 ### **...**

lcof2/剑指 Offer II 106. 二分图/Solution.cpp

@@ -0,0 +1,25 @@
+class Solution {
+public:
+    vector<int> p;
+
+    bool isBipartite(vector<vector<int>>& graph) {
+        int n = graph.size();
+        p.resize(n);
+        for (int i = 0; i < n; ++i) p[i] = i;
+        for (int u = 0; u < n; ++u)
+        {
+            auto g = graph[u];
+            for (int v : g)
+            {
+                if (find(u) == find(v)) return false;
+                p[find(v)] = find(g[0]);
+            }
+        }
+        return true;
+    }
+
+    int find(int x) {
+        if (p[x] != x) p[x] = find(p[x]);
+        return p[x];
+    }
+};

lcof2/剑指 Offer II 106. 二分图/Solution.go

@@ -0,0 +1,25 @@
+var p []int
+
+func isBipartite(graph [][]int) bool {
+	n := len(graph)
+	p = make([]int, n)
+	for i := 0; i < n; i++ {
+		p[i] = i
+	}
+	for u, g := range graph {
+		for _, v := range g {
+			if find(u) == find(v) {
+				return false
+			}
+			p[find(v)] = find(g[0])
+		}
+	}
+	return true
+}
+
+func find(x int) int {
+	if p[x] != x {
+		p[x] = find(p[x])
+	}
+	return p[x]
+}

lcof2/剑指 Offer II 106. 二分图/Solution.java

@@ -0,0 +1,28 @@
+class Solution {
+    private int[] p;
+
+    public boolean isBipartite(int[][] graph) {
+        int n = graph.length;
+        p = new int[n];
+        for (int i = 0; i < n; ++i) {
+            p[i] = i;
+        }
+        for (int u = 0; u < n; ++u) {
+            int[] g = graph[u];
+            for (int v : g) {
+                if (find(u) == find(v)) {
+                    return false;
+                }
+                p[find(v)] = find(g[0]);
+            }
+        }
+        return true;
+    }
+
+    private int find(int x) {
+        if (p[x] != x) {
+            p[x] = find(p[x]);
+        }
+        return p[x];
+    }
+}

lcof2/剑指 Offer II 106. 二分图/Solution.py

@@ -0,0 +1,16 @@
+class Solution:
+    def isBipartite(self, graph: List[List[int]]) -> bool:
+        n = len(graph)
+        p = list(range(n))
+
+        def find(x):
+            if p[x] != x:
+                p[x] = find(p[x])
+            return p[x]
+
+        for u, g in enumerate(graph):
+            for v in g:
+                if find(u) == find(v):
+                    return False
+                p[find(v)] = find(g[0])
+        return True