LC.P851[喧闹和富有]

方法一:拓扑排序

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class Solution {
int N = 510, M = N * N + 10;
int[] he = new int[N], e = new int[M], ne = new int[M];
int idx;

private void add(int a, int b) {
e[idx] = b;
ne[idx] = he[a];
he[a] = idx;
++idx;
}

public int[] loudAndRich(int[][] richer, int[] quiet) {
int n = quiet.length;
int[] in = new int[n];
Arrays.fill(he, -1);
for (int[] r : richer) {
int a = r[0], b = r[1];
add(a, b);
++in[b];
}
Deque<Integer> queue = new ArrayDeque<>();
int[] ans = new int[n];
for (int i = 0; i < n; i++) {
ans[i] = i;
if (in[i] == 0) queue.offer(i);
}
while (!queue.isEmpty()) {
int cur = queue.poll();
for (int i = he[cur]; i != -1; i = ne[i]) {
int j = e[i];
if (quiet[ans[cur]] < quiet[ans[j]]) ans[j] = ans[cur];
if (--in[j] == 0) queue.offer(j);
}
}
return ans;
}
}
  • 时间复杂度:$O(m + n)$,其中,$m$为richer的长度(边数),$n$为quiet的长度
  • 空间复杂度:$O(m + n)$