[LeetCode] 323. Number of Connected Components in an Undirected Graph 无向图中的连通区域的个数
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.
这道题和305. numbers of islands II 是一个思路,一个count初始化为n,union find每次有新的edge就union两个节点,如果两个节点(u, v)原来不在一个连通图里面就减少count并且连起来,如果原来就在一个图里面就不管。用一个索引array来做,union find优化就是加权了,每次把大的树的root当做parent,小的树的root作为child。
Java:
public class Solution {
public int countComponents(int n, int[][] edges) {
int count = n;
// array to store parent
init(n, edges);
for(int[] edge : edges) {
int root1 = find(edge[0]);
int root2 = find(edge[1]);
if(root1 != root2) {
union(root1, root2);
count--;
}
}
return count;
} int[] map;
private void init(int n, int[][] edges) {
map = new int[n];
for(int[] edge : edges) {
map[edge[0]] = edge[0];
map[edge[1]] = edge[1];
}
} private int find(int child) {
while(map[child] != child) child = map[child];
return child;
} private void union(int child, int parent) {
map[child] = parent;
}
}
Python:
# Time: O(nlog*n) ~= O(n), n is the length of the positions
# Space: O(n) class UnionFind(object):
def __init__(self, n):
self.set = range(n)
self.count = n def find_set(self, x):
if self.set[x] != x:
self.set[x] = self.find_set(self.set[x]) # path compression.
return self.set[x] def union_set(self, x, y):
x_root, y_root = map(self.find_set, (x, y))
if x_root != y_root:
self.set[min(x_root, y_root)] = max(x_root, y_root)
self.count -= 1 class Solution(object):
def countComponents(self, n, edges):
"""
:type n: int
:type edges: List[List[int]]
:rtype: int
"""
union_find = UnionFind(n)
for i, j in edges:
union_find.union_set(i, j)
return union_find.count
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