LeetCode - Is Graph Bipartite?

发布时间：2020-12-14 03:45:08 所属栏目：大数据来源：网络整理

导读：Given an undirected 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 n

Given an undirected 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],[1,2]]
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: [[1,2,1,2]]
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 subsets.

原来输入数组中的graph[i]，表示顶点i所有相邻的顶点，比如对于例子1来说，顶点0和顶点1，3相连，顶点1和顶点0，2相连，顶点2和结点1，3相连，顶点3和顶点0，2相连。这道题让我们验证给定的图是否是二分图，所谓二分图，就是可以将图中的所有顶点分成两个不相交的集合，使得同一个集合的顶点不相连。为了验证是否有这样的两个不相交的集合存在，我们采用一种很机智的染色法，大体上的思路是要将相连的两个顶点染成不同的颜色，一旦在染的过程中发现有两连的两个顶点已经被染成相同的颜色，说明不是二分图。这里我们使用两种颜色，分别用1和-1来表示，初始时每个顶点用0表示未染色，然后遍历每一个顶点，如果该顶点未被访问过，则调用递归函数，如果返回false，那么说明不是二分图，则直接返回false。如果循环退出后没有返回false，则返回true。在递归函数中，如果当前顶点已经染色，如果该顶点的颜色和将要染的颜色相同，则返回true，否则返回false。如果没被染色，则将当前顶点染色，然后再遍历与该顶点相连的所有的顶点，调用递归函数，如果返回false了，则当前递归函数的返回false，循环结束返回true，参见代码如下：

class Solution {
    public boolean isBipartite(int[][] graph) {
        if(graph == null){
            return false;
        }
        int v = graph.length;
        //if we only have 2 vertices,then it‘s a Bipartite
        if(v <= 2){
            return true;
        }
        int[] colors = new int[v];
        for(int i = 0; i < v; i++){
            if(colors[i] == 0 && !valid(graph,i,1,colors)){
                return false;
            }
        }
        return true;
    }
    
    private boolean valid (int[][] graph,int cur,int color,int[] colors){
        if(colors[cur] != 0){
            return colors[cur] == color;
        }
        colors[cur] = color;
        for(int i : graph[cur]){
            if(!valid(graph,color * -1,colors)){
                return false;
            }
        }
        return true;
    }
}

（编辑：李大同）

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