Question

547. Number of Provinces

中文文档

Description

There are n cities. Some of them are connected, while some are not. If city a is connected directly with city b, and city b is connected directly with city c, then city a is connected indirectly with city c.

A province is a group of directly or indirectly connected cities and no other cities outside of the group.

You are given an n x n matrix isConnected where isConnected[i][j] = 1 if the ith city and the jth city are directly connected, and isConnected[i][j] = 0 otherwise.

Return _the total number of provinces_.

 

Example 1:

Input: isConnected = [[1,1,0],[1,1,0],[0,0,1]]
Output: 2

Example 2:

Input: isConnected = [[1,0,0],[0,1,0],[0,0,1]]
Output: 3

 

Constraints:

• 1 <= n <= 200
• n == isConnected.length
• n == isConnected[i].length
• isConnected[i][j] is 1 or 0.
• isConnected[i][i] == 1
• isConnected[i][j] == isConnected[j][i]

Solutions

Solution 1

class Solution:
  def findCircleNum(self, isConnected: List[List[int]]) -> int:
    def dfs(i: int):
      vis[i] = True
      for j, x in enumerate(isConnected[i]):
        if not vis[j] and x:
          dfs(j)

    n = len(isConnected)
    vis = [False] * n
    ans = 0
    for i in range(n):
      if not vis[i]:
        dfs(i)
        ans += 1
    return ans

class Solution {
  private int[][] g;
  private boolean[] vis;

  public int findCircleNum(int[][] isConnected) {
    g = isConnected;
    int n = g.length;
    vis = new boolean[n];
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      if (!vis[i]) {
        dfs(i);
        ++ans;
      }
    }
    return ans;
  }

  private void dfs(int i) {
    vis[i] = true;
    for (int j = 0; j < g.length; ++j) {
      if (!vis[j] && g[i][j] == 1) {
        dfs(j);
      }
    }
  }
}

class Solution {
public:
  int findCircleNum(vector<vector<int>>& isConnected) {
    int n = isConnected.size();
    int ans = 0;
    bool vis[n];
    memset(vis, false, sizeof(vis));
    function<void(int)> dfs = [&](int i) {
      vis[i] = true;
      for (int j = 0; j < n; ++j) {
        if (!vis[j] && isConnected[i][j]) {
          dfs(j);
        }
      }
    };
    for (int i = 0; i < n; ++i) {
      if (!vis[i]) {
        dfs(i);
        ++ans;
      }
    }
    return ans;
  }
};

func findCircleNum(isConnected [][]int) (ans int) {
  n := len(isConnected)
  vis := make([]bool, n)
  var dfs func(int)
  dfs = func(i int) {
    vis[i] = true
    for j, x := range isConnected[i] {
      if !vis[j] && x == 1 {
        dfs(j)
      }
    }
  }
  for i, v := range vis {
    if !v {
      ans++
      dfs(i)
    }
  }
  return
}

function findCircleNum(isConnected: number[][]): number {
  const n = isConnected.length;
  const vis: boolean[] = new Array(n).fill(false);
  const dfs = (i: number) => {
    vis[i] = true;
    for (let j = 0; j < n; ++j) {
      if (!vis[j] && isConnected[i][j]) {
        dfs(j);
      }
    }
  };
  let ans = 0;
  for (let i = 0; i < n; ++i) {
    if (!vis[i]) {
      dfs(i);
      ++ans;
    }
  }
  return ans;
}

impl Solution {
  fn dfs(is_connected: &mut Vec<Vec<i32>>, vis: &mut Vec<bool>, i: usize) {
    vis[i] = true;
    for j in 0..is_connected.len() {
      if vis[j] || is_connected[i][j] == 0 {
        continue;
      }
      Self::dfs(is_connected, vis, j);
    }
  }

  pub fn find_circle_num(mut is_connected: Vec<Vec<i32>>) -> i32 {
    let n = is_connected.len();
    let mut vis = vec![false; n];
    let mut res = 0;
    for i in 0..n {
      if vis[i] {
        continue;
      }
      res += 1;
      Self::dfs(&mut is_connected, &mut vis, i);
    }
    res
  }
}

Solution 2

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

    n = len(isConnected)
    p = list(range(n))
    ans = n
    for i in range(n):
      for j in range(i + 1, n):
        if isConnected[i][j]:
          pa, pb = find(i), find(j)
          if pa != pb:
            p[pa] = pb
            ans -= 1
    return ans

class Solution {
  private int[] p;

  public int findCircleNum(int[][] isConnected) {
    int n = isConnected.length;
    p = new int[n];
    for (int i = 0; i < n; ++i) {
      p[i] = i;
    }
    int ans = n;
    for (int i = 0; i < n; ++i) {
      for (int j = i + 1; j < n; ++j) {
        if (isConnected[i][j] == 1) {
          int pa = find(i), pb = find(j);
          if (pa != pb) {
            p[pa] = pb;
            --ans;
          }
        }
      }
    }
    return ans;
  }

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

class Solution {
public:
  int findCircleNum(vector<vector<int>>& isConnected) {
    int n = isConnected.size();
    int p[n];
    iota(p, p + n, 0);
    function<int(int)> find = [&](int x) -> int {
      if (p[x] != x) {
        p[x] = find(p[x]);
      }
      return p[x];
    };
    int ans = n;
    for (int i = 0; i < n; ++i) {
      for (int j = i + 1; j < n; ++j) {
        if (isConnected[i][j]) {
          int pa = find(i), pb = find(j);
          if (pa != pb) {
            p[pa] = pb;
            --ans;
          }
        }
      }
    }
    return ans;
  }
};

func findCircleNum(isConnected [][]int) (ans int) {
  n := len(isConnected)
  p := make([]int, n)
  for i := range p {
    p[i] = i
  }
  var find func(x int) int
  find = func(x int) int {
    if p[x] != x {
      p[x] = find(p[x])
    }
    return p[x]
  }
  ans = n
  for i := 0; i < n; i++ {
    for j := 0; j < n; j++ {
      if isConnected[i][j] == 1 {
        pa, pb := find(i), find(j)
        if pa != pb {
          p[pa] = pb
          ans--
        }
      }
    }
  }
  return
}

function findCircleNum(isConnected: number[][]): number {
  const n = isConnected.length;
  const p: number[] = new Array(n);
  for (let i = 0; i < n; ++i) {
    p[i] = i;
  }
  const find = (x: number): number => {
    if (p[x] !== x) {
      p[x] = find(p[x]);
    }
    return p[x];
  };
  let ans = n;
  for (let i = 0; i < n; ++i) {
    for (let j = i + 1; j < n; ++j) {
      if (isConnected[i][j]) {
        const pa = find(i);
        const pb = find(j);
        if (pa !== pb) {
          p[pa] = pb;
          --ans;
        }
      }
    }
  }
  return ans;
}