Leetcode题解——DFS+染色问题

涉及到的题目：

886. 可能的二分法

785. 判断二分图

此类题目的关键在于构建图，然后将一条边上两个点着不同的颜色，当着色方案为2种颜色时，即为所说的二分图问题。

886. 可能的二分法

class Solution:
    def possibleBipartition(self, N: int, dislikes: List[List[int]]) -> bool:
        color = {}
        graph = collections.defaultdict(set)
        for i,j in dislikes:
            graph[i].add(j)
            graph[j].add(i)

        def dfs(i):
            nei = graph[i]
            for j in nei:
                if j not in color:
                    color[j] = color[i] ^ 1
                    if not dfs(j): return False
                elif color[j] == color[i]:
                    return False
            return True

        for i in range(1,N+1):
            if i not in color:
                color[i] = 0
                if not dfs(i): return False
        return True

785. 判断二分图

DFS版本：

class Solution(object):
    def isBipartite(self, graph):
        """
        :type graph: List[List[int]]
        :rtype: bool
        """
        n = len(graph)
        visited = [-1] * n
        for i in range(n):
            if visited[i] == -1:
                if not self.dfs(graph, i, 0, visited):
                    return False
        return True

    def dfs(self, graph, v, color, visited):
        visited[v] = color
        for i in graph[v]:
            if visited[i] == -1:
                if not self.dfs(graph, i, 1 - color, visited):
                    return False
            elif visited[i] == color:
                return False
        return True

BFS版本：

class Solution(object):
    def isBipartite(self, graph):
        n = len(graph)
        visited = [-1] * n
        for i in range(n):
            if visited[i] == -1:
                if not self.bfs(graph, i, 0, visited):
                    return False
        return True

    def bfs(self, graph, v, color, visited):
        visited[v], queue = color, [v]
        while queue:
            node = queue.pop(0)
            for i in graph[node]:
                if visited[i] == -1:
                    visited[i] = 1 - visited[node]
                    queue.append(i)
                elif visited[i] == visited[node]:
                    return False
        return True