发布于 2016-01-12 09:17:37 | 973 次阅读 | 评论: 0 | 来源: PHPERZ

这里有新鲜出炉的精品教程，程序狗速度看过来！

Leetcode 在线编程网站

leetcode 是一个美国的在线编程网站，上面主要收集了各大IT公司的笔试面试题，对于应届毕业生找工作是一个不可多得的好帮手。

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Number of Connected Components in an Undirected Graph

Given n nodes labeled from 0 to n - 1 and a list of undirected edges (each edge is a pair of nodes), write a function to find the number of connected components in an undirected graph.

Example 1:

    0          3
    |          |
    1 --- 2    4

Given n = 5 and edges = [[0, 1], [1, 2], [3, 4]], return 2.

Example 2:

    0           4
    |           |
    1 --- 2 --- 3

Given n = 5 and edges = [[0, 1], [1, 2], [2, 3], [3, 4]], return 1.

分析

典型且很基础的union find题。用一个数组记录各个数字的父节点，然后遍历图，对edge中两个端点做union。最后扫一遍数组，找到根节点个数即可。

time:

time: O(m*h), space: O(n), m表示edge的数量。

代码

public class Solution {
    public int countComponents(int n, int[][] edges) {
        int[] id = new int[n];
        
        // 初始化
        for (int i = 0; i < n; i++) {
            id[i] = i;
        }
        
        // union
        for (int[] edge : edges) {              
            int i = root(id, edge[0]);
            int j = root(id, edge[1]);
            id[i] = j;
        }
        
        // 统计根节点个数
        int count = 0;
        for (int i = 0; i < n; i++) {
            if (id[i] == i)
                count++;
        }
        return count;
    }
    
    // 找根节点
    public int root(int[] id, int i) {
        while (i != id[i]) {
            id[i] = id[id[i]];
            i = id[i];
        }
        return i;
    }
}