标签:
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
.
Note:
You can assume that no duplicate edges will appear in edges
. Since all edges are undirected, [0, 1]
is the same as [1, 0]
and thus will not appear together in edges
.
Analysis:
From here: https://discuss.leetcode.com/topic/32752/easiest-2ms-java-solution
This is 1D version of Number of Islands II. For more explanations, check out this 2D Solution.
n
points = n
islands = n
trees = n
roots.e[0]
or e[1]
belonging to.e[0]
and e[1]
are in same islands, do nothing.1
.50%
.Hope it helps!
Solution:
1 public class Solution { 2 public int countComponents(int n, int[][] edges) { 3 int[] roots = new int[n]; 4 for (int i=0;i<n;i++) roots[i] = i; 5 int num = n; 6 7 int len = edges.length; 8 for (int i=0;i<len;i++){ 9 int root1 = findRoot(roots,edges[i][0]); 10 int root2 = findRoot(roots,edges[i][1]); 11 12 if (root1!=root2){ 13 roots[root1] = root2; 14 num--; 15 } 16 } 17 18 return num; 19 } 20 21 public int findRoot(int[] roots, int node){ 22 while (roots[node]!=node){ 23 roots[node] = roots[roots[node]]; 24 node = roots[node]; 25 } 26 return node; 27 } 28 }
LeetCode-Number of Connected Components in an Undirected Graph
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原文地址:http://www.cnblogs.com/lishiblog/p/5764639.html