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剑指 Offer II 116. 朋友圈

题目描述

一个班上有 n 个同学，其中一些彼此是朋友，另一些不是。朋友关系是可以传递的，如果 a 与 b 直接是朋友，且 b 与 c 是直接朋友，那么 a 与 c 就是间接朋友。

定义 朋友圈 就是一组直接或者间接朋友的同学集合。

给定一个 n x n 的矩阵 isConnected 表示班上的朋友关系，其中 isConnected[i][j] = 1 表示第 i 个同学和第 j 个同学是直接朋友，而 isConnected[i][j] = 0 表示二人不是直接朋友。

返回矩阵中 朋友圈的数量。

 

示例 1：

输入：isConnected = [[1,1,0],[1,1,0],[0,0,1]]
输出：2

示例 2：

输入：isConnected = [[1,0,0],[0,1,0],[0,0,1]]
输出：3

 

提示：

  • 1 <= n <= 200
  • n == isConnected.length
  • n == isConnected[i].length
  • isConnected[i][j] 为 1 或 0
  • isConnected[i][i] == 1
  • isConnected[i][j] == isConnected[j][i]

 

注意：本题与主站 547 题相同： https://leetcode.cn/problems/number-of-provinces/

解法

方法一：深度优先搜索

判断城市之间是否属于同一个连通分量，最后连通分量的总数即为结果。

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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

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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);
            }
        }
    }
}

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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);
    }
};

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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
}

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class Solution {
    private var isConnected: [[Int]] = []
    private var visited: [Bool] = []
    private var n: Int = 0

    func findCircleNum(_ isConnected: [[Int]]) -> Int {
        self.isConnected = isConnected
        self.n = isConnected.count
        self.visited = [Bool](repeating: false, count: n)
        var numberOfCircles = 0

        for i in 0..<n {
            if !visited[i] {
                dfs(i)
                numberOfCircles += 1
            }
        }
        return numberOfCircles
    }

    private func dfs(_ i: Int) {
        visited[i] = true
        for j in 0..<n {
            if !visited[j] && isConnected[i][j] == 1 {
                dfs(j)
            }
        }
    }
}

方法二：并查集

模板 1——朴素并查集：

# 初始化，p存储每个点的父节点
p = list(range(n))


# 返回x的祖宗节点
def find(x):
    if p[x] != x:
        # 路径压缩
        p[x] = find(p[x])
    return p[x]


# 合并a和b所在的两个集合
p[find(a)] = find(b)

模板 2——维护 size 的并查集：

# 初始化，p存储每个点的父节点，size只有当节点是祖宗节点时才有意义，表示祖宗节点所在集合中，点的数量
p = list(range(n))
size = [1] * n


# 返回x的祖宗节点
def find(x):
    if p[x] != x:
        # 路径压缩
        p[x] = find(p[x])
    return p[x]


# 合并a和b所在的两个集合
if find(a) != find(b):
    size[find(b)] += size[find(a)]
    p[find(a)] = find(b)

模板 3——维护到祖宗节点距离的并查集：

# 初始化，p存储每个点的父节点，d[x]存储x到p[x]的距离
p = list(range(n))
d = [0] * n


# 返回x的祖宗节点
def find(x):
    if p[x] != x:
        t = find(p[x])
        d[x] += d[p[x]]
        p[x] = t
    return p[x]


# 合并a和b所在的两个集合
p[find(a)] = find(b)
d[find(a)] = distance

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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))

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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];
    }
}

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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];
    }
};

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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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