323. Number of Connected Components in an Undirected Graph按照线段添加的并查集

时间:2021-11-02 14:58:48

[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 and edges = [[0, 1], [1, 2], [3, 4]]

     0          3
| |
1 --- 2 4 Output: 2

Example 2:

Input: n = 5 and edges = [[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;
}
}