[抄题]:
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:
Input:n = 5
andedges = [[0, 1], [1, 2], [3, 4]]
0 3
| |
1 --- 2 4 Output: 2
Example 2:
Input:n = 5
andedges = [[0, 1], [1, 2], [2, 3], [3, 4]]
0 4
| |
1 --- 2 --- 3 Output: 1
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
不知道线段怎么加
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[一句话思路]:
线段也是由点构成的,分成两个点来加
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
每次更新的都是roots数组,把新的root指定给roots数组中的元素
//merge if neccessary
if (root1 != root0) {
roots[root1] = root0;
count--;
}
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
[复杂度]:Time complexity: O(1) Space complexity: O(n)
[算法思想:迭代/递归/分治/贪心]:递归
[关键模板化代码]:
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
[是否头一次写此类driver funcion的代码] :
[潜台词] :
class Solution {
public int countComponents(int n, int[][] edges) {
//use union find
//ini
int count = n;
int[] roots = new int[n]; //cc
if (n == 0 || edges == null) return 0; //initialization the roots as themselves
for (int i = 0; i < n; i++)
roots[i] = i; //add every edge
for (int[] edge : edges) {
int root0 = find(edge[0], roots);
int root1 = find(edge[1], roots); //merge if neccessary
if (root1 != root0) {
roots[root1] = root0;
count--;
}
} //return
return count; } public int find(int id, int[] roots) {
while (id != roots[id])
id = roots[roots[id]];
return id;
}
}