LeetCode 1059. All Paths from Source Lead to Destination
原题链接在这里:https://leetcode.com/problems/all-paths-from-source-lead-to-destination/
题目:
Given the edges
of a directed graph, and two nodes source
and destination
of this graph, determine whether or not all paths starting from source
eventually end at destination
, that is:
- At least one path exists from the
source
node to thedestination
node - If a path exists from the
source
node to a node with no outgoing edges, then that node is equal todestination
. - The number of possible paths from
source
todestination
is a finite number.
Return true
if and only if all roads from source
lead to destination
.
Example 1:
Input: n = 3, edges = [[0,1],[0,2]], source = 0, destination = 2
Output: false
Explanation: It is possible to reach and get stuck on both node 1 and node 2.
Example 2:
Input: n = 4, edges = [[0,1],[0,3],[1,2],[2,1]], source = 0, destination = 3
Output: false
Explanation: We have two possibilities: to end at node 3, or to loop over node 1 and node 2 indefinitely.
Example 3:
Input: n = 4, edges = [[0,1],[0,2],[1,3],[2,3]], source = 0, destination = 3
Output: true
Example 4:
Input: n = 3, edges = [[0,1],[1,1],[1,2]], source = 0, destination = 2
Output: false
Explanation: All paths from the source node end at the destination node, but there are an infinite number of paths, such as 0-1-2, 0-1-1-2, 0-1-1-1-2, 0-1-1-1-1-2, and so on.
Example 5:
Input: n = 2, edges = [[0,1],[1,1]], source = 0, destination = 1
Output: false
Explanation: There is infinite self-loop at destination node.
Note:
- The given graph may have self loops and parallel edges.
- The number of nodes
n
in the graph is between1
and10000
- The number of edges in the graph is between
0
and10000
0 <= edges.length <= 10000
edges[i].length == 2
0 <= source <= n - 1
0 <= destination <= n - 1
题解:
There are 2 cases it should return false.
case 1: it encounters a node that has no outgoing edges, but it is not destination.
case 2: it has cycle.
Otherwise, it returns true.
Could iterate graph with BFS. When indegree of a node becomes negative, then ther is cycle.
Time Complexity: O(n+e). e = edges.length.
Space: O(n+e).
AC Java:
class Solution {
public boolean leadsToDestination(int n, int[][] edges, int source, int destination) {
Set<Integer> [] graph = new Set[n]; for(int i = 0; i<n; i++){
graph[i] = new HashSet<Integer>();
} int [] inDegrees = new int[n];
for(int [] edge : edges){
graph[edge[0]].add(edge[1]);
inDegrees[edge[1]]++;
} LinkedList<Integer> que = new LinkedList<Integer>();
que.add(source); while(!que.isEmpty()){
int cur = que.poll();
if(graph[cur].size() == 0 && cur != destination){
return false;
} for(int nei : graph[cur]){
if(inDegrees[nei] < 0){
return false;
} inDegrees[nei]--; que.add(nei);
}
} return true;
}
}
Could iterate by DFS too.
If current node has been visited within current DFS, then there is cycle.
When traversing all the nodes, make current node as done.
Time Complexity: O(n+e).
Space: O(n+e).
AC Java:
class Solution {
public boolean leadsToDestination(int n, int[][] edges, int source, int destination) {
Set<Integer> [] graph = new Set[n];
for(int i = 0; i<n; i++){
graph[i] = new HashSet<Integer>();
} for(int [] edge : edges){
graph[edge[0]].add(edge[1]);
} return dfs(source, destination, graph, new int[n]);
} private boolean dfs(int cur, int destination, Set<Integer> [] graph, int [] visited){
if(visited[cur] != 0){
return visited[cur] == 2;
} if(graph[cur].size() == 0){
return cur == destination;
} visited[cur] = 1;
for(int nei : graph[cur]){
if(!dfs(nei, destination, graph, visited)){
return false;
}
} visited[cur] = 2;
return true;
}
}
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