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 ofedges. Each element ofedgesis a pair[u, v]that represents a directed edge connecting nodesuandv, whereuis a parent of childv.
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.
class Solution {
public int[] findRedundantDirectedConnection(int[][] edges) {
int[] res1 = {-1, -1};
int[] res2 = {-1, -1};
int n = edges.length;
int[] parents = new int[n + 1];
boolean hasDuplicateParents = false;
for (int[] edge : edges) {
int u = edge[0];
int v = edge[1];
// Find duplicate parents for node v
if (parents[v] > 0) {
res1[0] = parents[v];
res1[1] = v;
// res2 stores the extra edge that occurs LAST
res2[0] = u;
res2[1] = v;
// Remove the edge that cause duplicate parents
// If later on we still find cycle in the next step, we should return res1
edge[0] = -1;
edge[1] = -1;
hasDuplicateParents = true;
}
else {
parents[v] = u;
}
}
parents = new int[n + 1];
// Check if there is a cycle
for (int[] edge : edges) {
int u = edge[0];
int v = edge[1];
if (u < 0 || v < 0) {
continue;
}
parents[v] = u;
if (hasCycle(v, parents)) {
return hasDuplicateParents ? res1 : edge;
}
}
return res2;
}
public boolean hasCycle(int node, int[] parents) {
int parent = parents[node];
while (parent != 0) {
if (node == parent) {
return true;
}
parent = parents[parent];
}
return false;
}
}
(2) Union Find
思路: 这里唯一的区别就是用Union Find查找cycle
class Solution {
public int[] findRedundantDirectedConnection(int[][] edges) {
int[] res1 = new int[] {-1, -1};
int[] res2 = new int[] {-1, -1};
int n = edges.length;
int[] parent = new int[n + 1];
boolean hasDuplicateParents = false;
// Step 1: Check if there is duplicate parent
for (int[] edge : edges) {
// Duplicate parent is found
if (parent[edge[1]] > 0) {
res1 = new int[] {parent[edge[1]], edge[1]};
res2 = new int[] {edge[0], edge[1]};
hasDuplicateParents = true;
// Delete the current edge
// So that we can find cycle correctly in Step 2
edge[0] = -1;
edge[1] = -1;
}
else {
parent[edge[1]] = edge[0];
}
}
// Step 2: Use Union Find to find cycle, same as #684
// Here we don't consider the graph is directional
UnionFind uf = new UnionFind(n);
for (int[] edge : edges) {
// Skip deleted edge
if (edge[0] < 0 || edge[1] < 0) continue;
// If Cycle is found and no duplicate parent, return current edge
// Otherwise, return the edge that causes duplicate parent
if (!uf.union(edge[1], edge[0])) {
return hasDuplicateParents ? res1 : edge;
}
}
return res2;
}
class UnionFind {
int[] sets;
int[] size;
public UnionFind(int n) {
sets = new int[n + 1];
size = new int[n + 1];
for (int i = 1; i <= n; i++) {
sets[i] = i;
size[i] = 1;
}
}
public int find(int node) {
while (node != sets[node]) {
node = sets[node];
}
return node;
}
public boolean union(int i, int j) {
int node1 = find(i);
int node2 = find(j);
// Find Cycle
if (node1 == node2) {
return false;
}
if (size[node1] < size[node2]) {
sets[node1] = node2;
size[node2] += size[node1];
}
else {
sets[node2] = node1;
size[node1] += size[node2];
}
return true;
}
}
}
3. Time & Space Complexity
DFS: 时间复杂度O(n^2),n是node的个数，空间复杂度O(n)
Union Find: 时间复杂度O(nlogn), n为edge的个数，union()的时间复杂度是O(logn) 但实际上是接近常数的，所以总的时间复杂度可以看成是O(n), 空间复杂度O(n)