README_EN.md

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

Python3

DFS.

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

Union find.

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

        n = len(isConnected)
        p = list(range(n))
        for i in range(n):
            for j in range(i + 1, n):
                if isConnected[i][j]:
                    p[find(i)] = find(j)
        return sum(i == v for i, v in enumerate(p))

Java

DFS.

class Solution {
    private int[][] isConnected;
    private boolean[] vis;
    private int n;

    public int findCircleNum(int[][] isConnected) {
        n = isConnected.length;
        vis = new boolean[n];
        this.isConnected = isConnected;
        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 < n; ++j) {
            if (!vis[j] && isConnected[i][j] == 1) {
                dfs(j);
            }
        }
    }
}

Union find.

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;
        }
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if (isConnected[i][j] == 1) {
                    p[find(i)] = find(j);
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (i == p[i]) {
                ++ans;
            }
        }
        return ans;
    }

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

C++

DFS.

class Solution {
public:
    vector<vector<int>> isConnected;
    vector<bool> vis;
    int n;

    int findCircleNum(vector<vector<int>>& isConnected) {
        n = isConnected.size();
        vis.resize(n);
        this->isConnected = isConnected;
        int ans = 0;
        for (int i = 0; i < n; ++i)
        {
            if (!vis[i])
            {
                dfs(i);
                ++ans;
            }
        }
        return ans;
    }

    void dfs(int i) {
        vis[i] = true;
        for (int j = 0; j < n; ++j)
            if (!vis[j] && isConnected[i][j])
                dfs(j);
    }
};

Union find.

class Solution {
public:
    vector<int> p;

    int findCircleNum(vector<vector<int>>& isConnected) {
        int n = isConnected.size();
        p.resize(n);
        for (int i = 0; i < n; ++i) p[i] = i;
        for (int i = 0; i < n; ++i)
            for (int j = i + 1; j < n; ++j)
                if (isConnected[i][j])
                    p[find(i)] = find(j);
        int ans = 0;
        for (int i = 0; i < n; ++i)
            if (i == p[i])
                ++ans;
        return ans;

    }

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

Go

DFS.

func findCircleNum(isConnected [][]int) int {
    n := len(isConnected)
    vis := make([]bool, n)
    var dfs func(i int)
    dfs = func(i int) {
        vis[i] = true
        for j := 0; j < n; j++ {
            if !vis[j] && isConnected[i][j] == 1 {
                dfs(j)
            }
        }
    }
    ans := 0
    for i := 0; i < n; i++ {
        if !vis[i] {
            dfs(i)
            ans++
        }
    }
    return ans
}

Union find.

func findCircleNum(isConnected [][]int) 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]
	}
	for i := 0; i < n; i++ {
		for j := i + 1; j < n; j++ {
			if isConnected[i][j] == 1 {
				p[find(i)] = find(j)
			}
		}
	}
	ans := 0
	for i := range p {
		if p[i] == i {
			ans++
		}
	}
	return ans
}

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