785.Is Graph Bipartite?

Tags: Medium Depth-first Search Breadth-first Search Graph

Links: https://leetcode.com/problems/is-graph-bipartite/

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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], [1,3], [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 subsets.

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.
• The graph is undirected: if any element j is in graph[i], then i will be in graph[j].

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这种题的解法就是很典型的图染色问题，染色问题一般分为四色问题的染色，和二部图的染色。

用一个数组d去记录每个节点的颜色，因为只有两种颜色，不用考虑回退，所以只需要保证直接相邻的两个节点颜色不一致即可，初始化为-1，即未染色。对于每个未染色的节点，采用DFS进行深搜即可。

 class Solution {
     int n;
public:
    bool isBipartite(vector<vector<int>>& graph) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        n = graph.size();
        vector<int> d(n, -1);

        for (int i = 0; i < n; ++i) {
            if (d[i] == -1) {
                if (!DFS(i, 0, d, graph)) return false;
            }
        }

        return true;
    }

    bool DFS(int pos, int color, vector<int> & d, vector<vector<int>> & graph)
    {
        d[pos] = color ^ 1; //只需要用01染色

        for (int i = 0; i < graph[pos].size(); ++i) {
            if (d[graph[pos][i]] == -1) {
                if (!DFS(graph[pos][i], d[pos], d, graph)) return false;
            }
            else if (d[graph[pos][i]] != color) return false;
        }

        return true;
    }
};