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.
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.
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.
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.
Example 1:
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:
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:
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.
主要是分清楚三種情況，case1: 只有環，沒有node有兩個parent；case2：有一個node有兩個parent，刪除任意一個都行，所以我們刪除後出現的那個；case3:有一個node有兩個parent，必須刪除在環上的那個，所以我們先刪除後出現的那個，再進行union find，看看如果還是connected，那麼就返回第二個，否則返回第一個。
class Solution {
public int[] findRedundantDirectedConnection(int[][] edges) {
if (edges == null || edges.length == 0 || edges[0].length == 0) return new int[]{};
int n = edges.length;
int m = edges[0].length;
int[] union = new int[n+1];
int[] ans1 = new int[2];
int[] ans2 = new int[2];
for (int i = 0; i < n; i++) {
if (union[edges[i][1]] != 0) {
ans1[0] = union[edges[i][1]];
ans1[1] = edges[i][1];
ans2[0] = edges[i][0];
ans2[1] = edges[i][1];
} else {
union[edges[i][1]] = edges[i][0];
}
}
for (int i = 1; i <= n; i++) union[i] = i;
if (ans1[0] == 0) {
for (int i = 0; i < n; i++) {
int leftRoot = find(edges[i][0], union);
int rightRoot = find(edges[i][1], union);
if (leftRoot == rightRoot) return edges[i];
union[rightRoot] = leftRoot;
}
} else {
for (int i = 0; i < n; i++) {
if (edges[i][0] == ans2[0] && edges[i][1] == ans2[1]) continue;
int leftRoot = find(edges[i][0], union);
int rightRoot = find(edges[i][1], union);
union[rightRoot] = leftRoot;
}
}
int root = find(1, union);
for (int i = 2; i <=n; i++) if (find(i, union) != root) return ans1;
return ans2;
}
public int find(int n, int[] union) {
while (n != union[n]) n = union[n];
return n;
}
}