Question

323. 无向图中连通分量的数目

English Version

题目描述

你有一个包含 n 个节点的图。给定一个整数 n 和一个数组 edges ，其中 edges[i] = [ai, bi] 表示图中 ai 和 bi 之间有一条边。

返回 _图中已连接分量的数目_ 。

 

示例 1:

输入: n = 5, edges = [[0, 1], [1, 2], [3, 4]]
输出: 2

示例 2:

输入: n = 5, edges = [[0,1], [1,2], [2,3], [3,4]]
输出:  1

 

提示：

• 1 <= n <= 2000
• 1 <= edges.length <= 5000
• edges[i].length == 2
• 0 <= ai <= bi < n
• ai != bi
• edges 中不会出现重复的边

解法

方法一

class Solution:
  def countComponents(self, n: int, edges: List[List[int]]) -> int:
    def find(x):
      if p[x] != x:
        p[x] = find(p[x])
      return p[x]

    p = list(range(n))
    for a, b in edges:
      p[find(a)] = find(b)
    return sum(i == find(i) for i in range(n))

class Solution {
  private int[] p;

  public int countComponents(int n, int[][] edges) {
    p = new int[n];
    for (int i = 0; i < n; ++i) {
      p[i] = i;
    }
    for (int[] e : edges) {
      int a = e[0], b = e[1];
      p[find(a)] = find(b);
    }
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      if (i == find(i)) {
        ++ans;
      }
    }
    return ans;
  }

  private int find(int x) {
    if (p[x] != x) {
      p[x] = find(p[x]);
    }
    return p[x];
  }
}

class Solution {
public:
  int countComponents(int n, vector<vector<int>>& edges) {
    vector<int> p(n);
    iota(p.begin(), p.end(), 0);
    for (int i = 0; i < n; ++i) p[i] = i;
    function<int(int)> find = [&](int x) -> int {
      if (p[x] != x) p[x] = find(p[x]);
      return p[x];
    };
    for (auto& e : edges) {
      int a = e[0], b = e[1];
      p[find(a)] = find(b);
    }
    int ans = 0;
    for (int i = 0; i < n; ++i) ans += i == find(i);
    return ans;
  }
};

func countComponents(n int, edges [][]int) (ans int) {
  p := make([]int, n)
  for i := range p {
    p[i] = i
  }
  var find func(int) int
  find = func(x int) int {
    if p[x] != x {
      p[x] = find(p[x])
    }
    return p[x]
  }
  for _, e := range edges {
    a, b := e[0], e[1]
    p[find(a)] = find(b)
  }
  for i := 0; i < n; i++ {
    if i == find(i) {
      ans++
    }
  }
  return
}

/**
 * @param {number} n
 * @param {number[][]} edges
 * @return {number}
 */
var countComponents = function (n, edges) {
  let p = new Array(n);
  for (let i = 0; i < n; ++i) {
    p[i] = i;
  }
  function find(x) {
    if (p[x] != x) {
      p[x] = find(p[x]);
    }
    return p[x];
  }
  for (const [a, b] of edges) {
    p[find(a)] = find(b);
  }
  let ans = 0;
  for (let i = 0; i < n; ++i) {
    if (i == find(i)) {
      ++ans;
    }
  }
  return ans;
};