[LeetCode] 685. Redundant Connection II 冗余的链接之 II

2021年09月15日 阅读数:1
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In this problem, a rooted tree is a directed graph such that, there is exactly one node (the root) for which all other nodes are descendants of this node, plus every node has exactly one parent, except for the root node which has no parents.html

The given input is a directed graph that started as a rooted tree with N nodes (with distinct values 1, 2, ..., N), with one additional directed edge added. The added edge has two different vertices chosen from 1 to N, and was not an edge that already existed.java

The resulting graph is given as a 2D-array of edges. Each element of edges is a pair [u, v] that represents a directed edge connecting nodes u and v, where u is a parent of child v.node

Return an edge that can be removed so that the resulting graph is a rooted tree of N nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array.python

Example 1:post

Input: [[1,2], [1,3], [2,3]]
Output: [2,3]
Explanation: The given directed graph will be like this:
  1
 / \
v   v
2-->3 

Example 2:this

Input: [[1,2], [2,3], [3,4], [4,1], [1,5]]
Output: [4,1]
Explanation: The given directed graph will be like this:
5 <- 1 -> 2
     ^    |
     |    v
     4 <- 3

Note:url

  • The size of the input 2D-array will be between 3 and 1000.
  • Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.

684. Redundant Connection 的拓展,684题给的是无向图,只须要删掉组成环的最后一条边便可,检测环就好了。这题给的是有向图,就复杂多了,有多种状况存在,好比例子1就是无环,可是有入度为2的结点3。再好比例子2是有环,可是没有入度为2的结点。还有一种状况例子没有给出,就是既有环,又有入度为2的结点。code

解法:Union findorm

There are two cases for the tree structure to be invalid.
1) A node having two parents;
including corner case: e.g. [[4,2],[1,5],[5,2],[5,3],[2,4]]
2) A circle existshtm

If we can remove exactly 1 edge to achieve the tree structure, a single node can have at most two parents. 

1) Check whether there is a node having two parents.
If so, store them as candidates A and B, and set the second edge invalid.
2) Perform normal union find.

If the tree is now valid
    simply return candidate B
else if candidates not existing
    we find a circle, return current edge;
else
    remove candidate A instead of B.

Java:

class Solution {
    public int[] findRedundantDirectedConnection(int[][] edges) {
        int[] can1 = {-1, -1};
        int[] can2 = {-1, -1};
        int[] parent = new int[edges.length + 1];
        for (int i = 0; i < edges.length; i++) {
            if (parent[edges[i][1]] == 0) {
                parent[edges[i][1]] = edges[i][0];
            } else {
                can2 = new int[] {edges[i][0], edges[i][1]};
                can1 = new int[] {parent[edges[i][1]], edges[i][1]};
                edges[i][1] = 0;
            }
        }
        for (int i = 0; i < edges.length; i++) {
            parent[i] = i;
        }
        for (int i = 0; i < edges.length; i++) {
            if (edges[i][1] == 0) {
                continue;
            }
            int child = edges[i][1], father = edges[i][0];
            if (root(parent, father) == child) {
                if (can1[0] == -1) {
                    return edges[i];
                }
                return can1;
            }
            parent[child] = father;
        }
        return can2;
    }
    
    int root(int[] parent, int i) {
        while (i != parent[i]) {
            parent[i] = parent[parent[i]];
            i = parent[i];
        }   
        return i;
    }
}

Python:

# Time:  O(nlog*n) ~= O(n), n is the length of the positions
# Space: O(n)
class UnionFind(object):
    def __init__(self, n):
        self.set = range(n)
        self.count = n

    def find_set(self, x):
        if self.set[x] != x:
            self.set[x] = self.find_set(self.set[x])  # path compression.
        return self.set[x]

    def union_set(self, x, y):
        x_root, y_root = map(self.find_set, (x, y))
        if x_root == y_root or \
           y != y_root:  # already has a father
            return False
        self.set[y_root] = x_root
        self.count -= 1
        return True


class Solution(object):
    def findRedundantDirectedConnection(self, edges):
        """
        :type edges: List[List[int]]
        :rtype: List[int]
        """
        union_find = UnionFind(len(edges)+1)
        for edge in edges:
            if not union_find.union_set(*edge):
                return edge
        return []  

C++:

class Solution {
public:
    vector<int> findRedundantDirectedConnection(vector<vector<int>>& edges) {
        int n = edges.size();
        vector<int> parent(n+1, 0), candA, candB;
        // step 1, check whether there is a node with two parents
        for (auto &edge:edges) {
            if (parent[edge[1]] == 0)
                parent[edge[1]] = edge[0]; 
            else {
                candA = {parent[edge[1]], edge[1]};
                candB = edge;
                edge[1] = 0;
            }
        } 
        // step 2, union find
        for (int i = 1; i <= n; i++) parent[i] = i;
        for (auto &edge:edges) {
            if (edge[1] == 0) continue;
            int u = edge[0], v = edge[1], pu = root(parent, u);
            // Now every node only has 1 parent, so root of v is implicitly v
            if (pu == v) {
                if (candA.empty()) return edge;
                return candA;
            }
            parent[v] = pu;
        }
        return candB;
    }
private:
    int root(vector<int>& parent, int k) {
        if (parent[k] != k) 
            parent[k] = root(parent, parent[k]);
        return parent[k];
    }
};

 

 

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