每日算法系列【LeetCode 685】冗余连接 II

题目描述

示例1

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

示例2

输入：
[[1,2], [2,3], [3,4], [4,1], [1,5]]
输出：
[4,1]
解释：
5 <- 1 -> 2
     ^    |
     |    v
     4 <- 3

提示

代码

c++

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

    vector<int> findRedundantDirectedConnection(vector<vector<int>>& edges) {
        n = edges.size();
        memset(degree, 0, sizeof degree);
        for (auto e : edges) ++degree[e[1]];
        for (int i = n-1; i >= 0; --i) {
            if (degree[edges[i][1]] == 2 && !wrongEdge(edges, i).size()) {
                return edges[i];
            }
        }
        return wrongEdge(edges, n);
    }

    vector<int> wrongEdge(vector<vector<int>>& edges, int except) {
        init();
        for (int i = 0; i < n; ++i) {
            if (i == except) continue;
            int u = edges[i][0], v = edges[i][1];
            if (same(u, v)) return edges[i];
            join(u, v);
        }
        return {};
    }

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

    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;
        f[v] = u;
    }

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

python

class Solution:
    def findRedundantDirectedConnection(self, edges: List[List[int]]) -> List[int]:
        self.n = len(edges)
        degree = [0] * (self.n+1)
        for u, v in edges:
            degree[v] += 1
        for u, v in edges[::-1]:
            if degree[v] == 2 and len(self.wrongEdge(edges, [u, v])) == 0:
                return [u, v]
        return self.wrongEdge(edges, [])
        

    def wrongEdge(self, edges, ex):
        self.f = [i for i in range(self.n+1)]
        for u, v in edges:
            if [u, v] == ex:
                continue
            if self.same(u, v):
                return [u, v]
            self.join(u, v)
        return []
    
    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
        self.f[v] = u
    
    def same(self, u, v):
        u, v = self.find(u), self.find(v)
        return u == v