万灵之树
难度:
标签:
题目描述
English description is not available for the problem. Please switch to Chinese.
代码结果
运行时间: 1843 ms, 内存: 358.3 MB
/*
* 思路:
* 1. 使用 Java Stream 生成所有可能的二叉树结构。
* 2. 使用 Java Stream 生成所有可能的宝石排列。
* 3. 对每种组合模拟能量流动过程,记录形成的数字。
* 4. 检查生成的数字是否满足 num % p == target。
*/
import java.util.*;
import java.util.stream.Collectors;
import java.util.stream.IntStream;
public class SolutionStream {
public long countValidTrees(int[] gem, int p, int target) {
List<TreeNode> trees = generateTrees(gem.length);
return trees.stream().flatMap(tree ->
permute(gem).stream().filter(perm -> isValidTree(tree, perm, p, target))
).count();
}
private List<TreeNode> generateTrees(int n) {
// 生成所有可能的二叉树结构
// 此处省略具体实现
return new ArrayList<>();
}
private List<List<Integer>> permute(int[] nums) {
if (nums.length == 0) return Collections.singletonList(Collections.emptyList());
return IntStream.range(0, nums.length).boxed().flatMap(i ->
permute(IntStream.concat(Arrays.stream(nums, 0, i), Arrays.stream(nums, i + 1, nums.length)).toArray())
.stream().map(tail -> {
List<Integer> perm = new ArrayList<>();
perm.add(nums[i]);
perm.addAll(tail);
return perm;
})
).collect(Collectors.toList());
}
private boolean isValidTree(TreeNode tree, List<Integer> gem, int p, int target) {
// 模拟能量流动过程并检查结果
// 此处省略具体实现
return false;
}
static class TreeNode {
TreeNode left;
TreeNode right;
int value;
}
}
解释
方法:
This solution utilizes dynamic programming and bitmasking to generate all possible binary tree structures using the gems as leaves and then computes the numeric value generated by these trees. Special care is taken with mod arithmetic to handle the large numbers produced during the calculation. The core idea is to recursively calculate potential results for each subset of gems represented as bit masks, and combine these results according to the binary tree structure to check if any of these results modulo p equals the target. The bitmask represents which gems are included in a subtree, and recursive calls help construct the left and right subtrees, with memorization (caching) used to avoid redundant calculations.
时间复杂度:
O(2^n * n)
空间复杂度:
O(2^n)
代码细节讲解
🦆
在这个解决方案中,为什么选择使用位掩码表示宝石的子集?这种表示方法有什么特别的优势吗?
▷🦆
递归函数`find_all`中使用了哪些基本情况,并且如何处理只有一个宝石的情况?
▷🦆
在处理模运算时,为什么需要计算`10^(p-2) % p`,这个计算结果是如何在后续的函数中使用的?
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