[LeetCode]Is Graph Bipartite?

题目描述:

LeetCode 785. Is Graph Bipartite?

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: 
The graph looks like this:
0----1
|    |
|    |
3----2
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: 
The graph looks like this:
0----1
| \  |
|  \ |
3----2
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.

题目大意:

以邻接表形式给定无向图graph,判断是否为二分图

解题思路:

染色 + DFS

遍历顶点,若顶点未被染色,则染色为红色(0),并将其相邻节点染成黑色(1),递归执行;若遇到冲突,则返回False

Python代码:

class Solution(object):
    def isBipartite(self, graph):
        """
        :type graph: List[List[int]]
        :rtype: bool
        """
        edges = collections.defaultdict(set)
        for idx, points in enumerate(graph):
            for p in points: edges[idx].add(p)
        colors = {}
        def dfs(v, color):
            ocolor = 1 - color
            for p in edges[v]:
                if p not in colors:
                    colors[p] = ocolor
                    if not dfs(p, ocolor):
                        return False
                elif colors[p] != ocolor:
                    return False
            return True
        for k in edges:
            if k in colors: continue
            colors[k] = 0
            if not dfs(k, 0): return False
        return True

错误解法:

将邻接表转化为邻接矩阵

然后判断每一组顶点集互相之间是否有边相连,若存在则返回False

返回True

Python代码(Wrong Answer):

class Solution(object):
    def isBipartite(self, graph):
        """
        :type graph: List[List[int]]
        :rtype: bool
        """
        edges = collections.defaultdict(set)
        for idx, points in enumerate(graph):
            for p in points: edges[idx].add(p)
        for points in graph:
            for x in range(len(points)):
                for y in range(x + 1, len(points)):
                    if points[y] in edges[points[x]]:
                        return False
        return True

 

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评论
  1. Joy Joy 发布于 2018年10月8日 14:48 #

    [[4,1],[0,2],[1,3],[2,4],[3,0]] 这个case您的代码通不过诶。。。

  2. 在线疯狂 在线疯狂 发布于 2018年12月2日 15:03 #

    感谢指正,已经更新。

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