演算法

1function bfs(adj: number[][], start: number): number[] {
2  const seen = new Set([start]);
3  const queue = [start];
4  const order: number[] = [];
5  for (let head = 0; head < queue.length; head++) {
6    const u = queue[head];
7    order.push(u);
8    for (const v of adj[u]) {
9      if (seen.has(v)) continue;
10      seen.add(v);
11      queue.push(v);
12    }
13  }
14  return order;
15}
16 
17function dfs(adj: number[][], start: number): number[] {
18  const visited = new Set<number>();
19  const stack = [start];
20  const order: number[] = [];
21  while (stack.length > 0) {
22    const u = stack.pop()!;
23    if (visited.has(u)) continue;
24    visited.add(u);
25    order.push(u);
26    for (let i = adj[u].length - 1; i >= 0; i--) {
27      if (!visited.has(adj[u][i])) stack.push(adj[u][i]);
28    }
29  }
30  return order;
31}