[LeetCode] Is Graph Bipartite? 是二分图么？

时间：2018-03-07 00:59:27 阅读：275 评论：0 收藏：0 [点我收藏+]

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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].

这道题博主在最开始做的时候，看了半天，愣是没弄懂输出数据的意思，博主开始以为给的是边，后来发现跟图对应不上，就懵逼了，后来是通过研究论坛上大神们的解法，才总算搞懂了题目的意思，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

解法一：

class Solution {
public:
    bool isBipartite(vector<vector<int>>& graph) {
        vector<int> colors(graph.size());
        for (int i = 0; i < graph.size(); ++i) {
            if (colors[i] == 0 && !valid(graph, 1, i, colors)) {
                return false;
            }
        }
        return true;
    }
    bool valid(vector<vector<int>>& graph, int color, int cur, vector<int>& colors) {
        if (colors[cur] != 0) return colors[cur] == color;
        colors[cur] = color;
        for (int i : graph[cur]) {
            if (!valid(graph, -1 * color, i, colors)) {
                return false;
            }
        }
        return true;
    }
};

解法二：

class Solution {
public:
    bool isBipartite(vector<vector<int>>& graph) {
        vector<int> colors(graph.size());
        for (int i = 0; i < graph.size(); ++i) {
            if (colors[i] == 0) colors[i] = 1;
            for (auto a : graph[i]) {
                if (colors[a] == colors[i]) return false;
                colrs[a] = -1 * colors[a];
            }
        }
        return true;
    }
};