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# 743.Network-Delay-Time

## 题目描述

There are N network nodes, labelled 1 to N.
Given times, a list of travel times as directed edges times[i] = (u, v, w), where u is the source node, v is the target node, and w is the time it takes for a signal to travel from source to target.
Now, we send a signal from a certain node K. How long will it take for all nodes to receive the signal? If it is impossible, return -1.
Example 1:
Input: times = [[2,1,1],[2,3,1],[3,4,1]], N = 4, K = 2
Output: 2
Note:
N will be in the range [1, 100].
K will be in the range [1, N].
The length of times will be in the range [1, 6000].
All edges times[i] = (u, v, w) will have 1 <= u, v <= N and 0 <= w <= 100.

## 代码

Time: O(N^N + E log E) && Space: O(N + E) where E is the length of `times`.
class Solution {
Map<Integer, Integer> dist;
public int networkDelayTime(int[][] times, int N, int K) {
Map<Integer, List<int[]>> graph = new HashMap();
for (int[] edge: times) {
if (!graph.containsKey(edge))
graph.put(edge, new ArrayList<int[]>());
}
for (int node: graph.keySet()) {
Collections.sort(graph.get(node), (a, b) -> a - b);
}
dist = new HashMap();
for (int node = 1; node <= N; ++node)
dist.put(node, Integer.MAX_VALUE);
dfs(graph, K, 0);
int ans = 0;
for (int cand: dist.values()) {
if (cand == Integer.MAX_VALUE) return -1;
ans = Math.max(ans, cand);
}
return ans;
}
public void dfs(Map<Integer, List<int[]>> graph, int node, int elapsed) {
if (elapsed >= dist.get(node)) return;
dist.put(node, elapsed);
if (graph.containsKey(node))
for (int[] info: graph.get(node))
dfs(graph, info, elapsed + info);
}
}