LWC 72: 785. Is Graph Bipartite?

Posted 2022-12-10 Demon的黑与白

LWC 72: 785. Is Graph Bipartite?

传送门：785. Is Graph Bipartite?

Problem:

Given a graph, return true if and only if it is bipartite.

Recall that a graph is bipartite if we can split it’s set of nodes into two independent subsets A and B such that every edge in the graph has one node in A and another node in B.

The graph is given in the following form: graph[i] is a list of indexes j for which the edge between nodes i and j exists. Each node is an integer between 0 and graph.length - 1. There are no self edges or parallel edges: graph[i] does not contain i, and it doesn’t contain any element twice.

Example 1:

Input: [[1,3], [0,2], [1,3], [0,2]]
Output: true
Explanation:
We can divide the vertices into two groups: 0, 2 and 1, 3.

Example 2:

Input: [[1,2,3], [0,2], [0,1,3], [0,2]]
Output: false
Explanation:
We cannot find a way to divide the set of nodes into two independent ubsets.

Note:

• graph will have length in range [1, 100].
• graph[i] will contain integers in range [0, graph.length - 1].
• graph[i] will not contain i or duplicate values.

思路：
二分图满足存在子集A和子集B，使得从A出发的边只能连接到B，B出发的边只能连接到A。换句话说，A出发的边不能连到A中的元素。B同理，采用染色法，因为相连的边一定是互异的，如果在不断染色的过程当中，出现冲突的情况，即返回false，全部染色完毕未发现冲突则返回true。

Java版本：

    boolean dfs(int[][] G, int[] color, int v, int c) 
        color[v] = c;
        for (int i = 0; i < color.length; ++i) 
            if (G[v][i] != 0) 
                if (color[i] != 0 && color[i] == c) return false;
                if (color[i] == 0 && !dfs(G, color, i, -c)) return false;
            
        
        return true;
    

    public boolean isBipartite(int[][] graph) 
        int n = graph.length;
        int[] color = new int[n];
        int[][] G = new int[n][n];
        for (int i = 0; i < n; ++i) 
            for (int j = 0; j < graph[i].length; ++j) 
                G[i][graph[i][j]] = 1;
            
        

        for (int i = 0; i < n; ++i) 
            if (color[i] == 0) 
                if (!dfs(G, color, i, 1)) return false;
            
        
        return true;
    

Python版本：

class Solution(object):
    def isBipartite(self, graph):
        """
        :type graph: List[List[int]]
        :rtype: bool
        """
        color = [0] * len(graph)
        return all(self.dfs(graph, color, v, 1) for v in range(len(graph)) if color[v] == 0)

    def dfs(self, graph, color, v, c):
        color[v] = c
        for to in graph[v]:
            if color[to] != 0 and color[to] == c: return False
            if color[to] == 0 and not self.dfs(graph, color, to, -c): return False
        return True