可能的二分法 -- 二分图判定【DFS、BFS分别实现】

python 复制代码
class PossibleBipartition:
    """
    可能的二分法
    「其实考察的就是二分图的判定」
    用dfs和bfs 两种方法分别实现
    https://leetcode.cn/problems/possible-bipartition/
    """
    def __init__(self):
        self.success = True
        self.color = []
        self.visited = []

    def dfs(self, n, dislikes):
        """
        DFS递归实现
        :param n: 
        :param dislikes:
        :return:
        """
        # 图节点编号为 1...n
        self.color = [False] * (n+1)
        self.visited = [False] * (n+1)
        graph = self.buildgraph(n, dislikes)

        # 因为图不一定是联通的，可能存在多个子图
        # 所以要把每个节点都作为起点进行一次遍历
        # 如果发现任何一个子图不是二分图，整幅图都不是二分图
        for v in range(1, n+1):
            if not self.visited[v]:
                self.dfs_traverse(graph, v)

        return self.success

    def buildgraph(self, n, dislikes):
        graph = [[] for _ in range(n+1)]

        for edge in dislikes:
            v = edge[1]
            w = edge[0]
            # 无向图相当于双向图
            graph[v].append(w)
            graph[w].append(v)

        return graph

    def dfs_traverse(self, graph, v):
        if not self.success:
            return

        self.visited[v] = True
        for w in graph[v]:
            if not self.visited[w]:
                self.color[w] = not self.color[v]
                self.dfs_traverse(graph, w)
            else:
                if self.color[v] == self.color[w]:
                    self.success = False
                    return

    def bfs(self, n, dislikes):
        """
        BFS实现，用队列替代递归调用
        :param n:
        :param dislikes:
        :return:
        """
        # 图节点编号为 1...n
        self.color = [False] * (n + 1)
        self.visited = [False] * (n + 1)
        graph = self.buildgraph(n, dislikes)

        # 因为图不一定是联通的，可能存在多个子图
        # 所以要把每个节点都作为起点进行一次遍历
        # 如果发现任何一个子图不是二分图，整幅图都不是二分图
        for v in range(1, n + 1):
            if not self.visited[v]:
                self.bfs_traverse(graph, v)

        return self.success

    def bfs_traverse(self, graph, start):
        # 节点队列
        queue = []
        self.visited[start] = True
        queue.append(start)

        while queue and self.success:
            v = queue.pop(0)
            # 从节点 v 向所有相邻节点扩散
            for w in graph[v]:
                if not self.visited[w]:
                    # 相邻节点w没有被访问过
                    # 那么应该给节点w涂上和节点v不同的颜⾊
                    self.color[w] = not self.color[v]
                    # 标记 w 节点，并放⼊队列
                    self.visited[w] = True
                    queue.append(w)
                else:
                    if self.color[v] == self.color[w]:
                        self.success = False
                        return