算法分析 | leetcode | 785. Is Graph Bipartite?

算法分析 | leetcode | 785. Is Graph Bipartite?

一、问题描述

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].

 

二、解法:

这道题我在最开始做的时候,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。

三、代码:

class Solution {
public:
    bool isBipartite(vector>& graph) {
        vector 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>& graph, int color, int cur, vector& 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;
    }
};

 

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