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); } }
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