[LeetCode] 1059. All Paths from Source Lead to Destination
作者:互联网
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
从始点到终点的所有路径。
给定有向图的边 edges,以及该图的始点 source 和目标终点 destination,确定从始点 source 出发的所有路径是否最终结束于目标终点 destination,即:
- 从始点 source 到目标终点 destination 存在至少一条路径
- 如果存在从始点 source 到没有出边的节点的路径,则该节点就是路径终点。
- 从始点source到目标终点 destination 可能路径数是有限数字
- 当从始点 source 出发的所有路径都可以到达目标终点 destination 时返回 true,否则返回 false。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/all-paths-from-source-lead-to-destination
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这是一道有向图的题。题目的意思解释的很清楚,给了起点,终点,所有的边,请你判断是不是所有从起点开始的边都能带你去到终点。
遇到图的题,绝大部分还是用DFS去遍历,得到答案。这个题需要注意的点是
- 当遇到一个终点(这个点没有next节点),判断这个点是不是destination
- 判断图中是否有环 - 染色法
思路不难想,代码的实现需要多练,面试才会写的6。
时间O(V + E)
空间O(n)
Java实现 - hashmap建图
1 class Solution { 2 public boolean leadsToDestination(int n, int[][] edges, int source, int destination) { 3 // build the graph 4 HashMap<Integer, List<Integer>> graph = new HashMap<>(); 5 for (int[] edge : edges) { 6 graph.putIfAbsent(edge[0], new ArrayList<>()); 7 graph.get(edge[0]).add(edge[1]); 8 } 9 return helper(graph, new HashSet<>(), source, destination); 10 } 11 12 private boolean helper(Map<Integer, List<Integer>> graph, Set<Integer> visited, int cur, int end) { 13 // base case 14 if (!graph.containsKey(cur)) { 15 return cur == end; 16 } 17 visited.add(cur); 18 for (int neighbor : graph.get(cur)) { 19 if (visited.contains(neighbor) || !helper(graph, visited, neighbor, end)) { 20 return false; 21 } 22 } 23 visited.remove(cur); 24 return true; 25 } 26 }View Code
Java实现 - list建图
1 class Solution { 2 public boolean leadsToDestination(int n, int[][] edges, int source, int destination) { 3 // corner case 4 if (edges == null || edges.length == 0) { 5 return true; 6 } 7 8 // normal case 9 List<Integer>[] g = new List[n]; 10 int[] colors = new int[n]; 11 buildGraph(g, edges); 12 return dfs(g, source, destination, colors); 13 } 14 15 // s = source, d = destination 16 private boolean dfs(List<Integer>[] g, int s, int d, int[] colors) { 17 // base case 18 if (g[s] == null || g[s].size() == 0) { 19 return s == d; 20 } 21 colors[s] = 1; 22 for (int next : g[s]) { 23 if (colors[next] == 1) { 24 return false; 25 } 26 if (colors[next] == 0 && !dfs(g, next, d, colors)) { 27 return false; 28 } 29 colors[s] = 2; 30 } 31 return true; 32 } 33 34 private void buildGraph(List<Integer>[] g, int[][] edges) { 35 for (int[] e : edges) { 36 int from = e[0]; 37 int to = e[1]; 38 if (g[from] == null) { 39 g[from] = new LinkedList<>(); 40 } 41 g[from].add(to); 42 } 43 } 44 }View Code
标签:node,Paths,Lead,int,1059,destination,source,edges,graph 来源: https://www.cnblogs.com/cnoodle/p/13375378.html