LC.P2368[受限条件下可到达节点的数目]

方法一：DFS

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class Solution {

    List<Integer>[] g;
    int ans;
    Set<Integer> set;

    public int reachableNodes(int n, int[][] edges, int[] restricted) {
        g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (int[] e : edges) {
            int a = e[0], b = e[1];
            g[a].add(b);
            g[b].add(a);
        }
        set = new HashSet<>();
        for (int e : restricted) {
            set.add(e);
        }
        dfs(0, -1);
        return ans;
    }

    private void dfs(int x, int pa) {
        ++ans;
        for (int y : g[x]) {
            if (y != pa) {
                if (set.contains(y)) continue;
                dfs(y, x);
            }
        }
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$

方法二：优化

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class Solution {

    List<Integer>[] g;

    public int reachableNodes(int n, int[][] edges, int[] restricted) {
        boolean[] isRestricted = new boolean[n];
        for (int x : restricted) {
            isRestricted[x] = true;
        }
        g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (int[] e : edges) {
            int a = e[0], b = e[1];
            // 都不受限才连边
            if (!isRestricted[a] && !isRestricted[b]) {
                g[a].add(b);
                g[b].add(a);
            }
        }
        return dfs(0, -1);
    }

    private int dfs(int x, int pa) {
        int ans = 1;
        for (int y : g[x]) {
            if (y != pa) {
                ans += dfs(y, x);
            }
        }
        return ans;
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$