每日算法系列【LeetCode 684】冗余连接

题目描述

在本问题中, 树指的是一个连通且无环的无向图。

示例1

输入：
[[1,2], [1,3], [2,3]]
输出：
[2,3]
解释：
  1
 / \
2 - 3

示例2

输入：
[[1,2], [2,3], [3,4], [1,4], [1,5]]
输出：
[1,4]
解释：
5 - 1 - 2
    |   |
    4 - 3

提示

代码

c++

class Solution {
public:
    static const int N = 1010;
    int f[N], rank[N];

    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        init();
        for (auto e : edges) {
            int u = e[0], v = e[1];
            if (same(u, v)) return {u, v};
            else join(u, v);
        }
        return {-1, -1};
    }

    void init() {
        for (int i = 0; i < N; ++i) {
            f[i] = i;
            rank[i] = 1;
        }
    }

    int find(int u) {
        return u==f[u] ? u : f[u]=find(f[u]);
    }

    void join(int u, int v) {
        u = find(u);
        v = find(v);
        if (u == v) return;
        if (rank[u] < rank[v]) {
            f[u] = v;
        } else {
            f[v] = u;
            if (rank[u] == rank[v]) {
                rank[u]++;
            }
        }
    }

    bool same(int u, int v) {
        u = find(u);
        v = find(v);
        return u == v;
    }
};

python

class Solution:
    def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
        n = len(edges)
        self.f = [i for i in range(n+1)]
        self.rank = [1] * (n+1)
        for [u, v] in edges:
            if self.same(u, v):
                return [u, v]
            else:
                self.join(u, v)
    
    def find(self, u):
        if u == self.f[u]:
            return u
        self.f[u] = self.find(self.f[u])
        return self.f[u]

    def join(self, u, v):
        u, v = self.find(u), self.find(v)
        if u == v:
            return
        if self.rank[u] < self.rank[v]:
            self.f[u] = v
        else:
            self.f[v] = u
            if self.rank[u] == self.rank[v]:
                self.rank[u] += 1
    
    def same(self, u, v):
        u, v = self.find(u), self.find(v)
        return u == v