华为8.19笔试第二题java的思路，求指点
看了题以后试了下java，求大佬指点。
除了各种边界的检查之外，大致思路是：
1.当前深度d可以摆放的node的最大值由上层node数量*2得到
2.当前深度数量为n的node的可摆放的种类可以简化为：n个苹果放入m个盘子里，每个盘子只能最多放一个苹果
3.把每层的可以摆放node的可能性相乘得到最后结果
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
public class Main2 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int nodeSize = n;
// save <depth, number of nodes at its depth> to map
Map<Integer, Integer> map = new HashMap<>();
int maxDepth = Integer.MIN_VALUE;
int mod = 1000000007;
while (nodeSize-- != 0) {
int node = sc.nextInt();
maxDepth = Math.max(maxDepth, node);
if (map.containsKey(node)) {
map.put(node, map.get(node) + 1);
} else {
map.put(node, 1);
}
}
if (!map.containsKey(0) || map.get(0) != 1) {
System.out.println("0");
return;
}
int d = 1;
int result = 1;
while (map.containsKey(d)) {
int curr = map.get(d);
int last = map.get(d - 1) * 2;
if (curr > last) {
System.out.println("0");
return;
}
result = (result*combination(curr,last))%mod;
d++;
}
System.out.println(result);
}
// m apple put into n plate, each plate can have max 1 apple
private static int combination(int curr, int last) {
if(curr == last || curr==0)
return 1;
return combination(curr-1,last-1) + combination(curr,last-1);
}
} 