[LeetCode每日一题]96. Unique Binary Search Trees
2020/3/6小于 1 分钟
题目如下:
Given n, how many structurally unique BST's (binary search trees) that store values 1 ... n?
Example:
Input: 3
Output: 5
Explanation:
Given n = 3, there are a total of 5 unique BST's:
1 3 3 2 1
\ / / / \ \
3 2 1 1 3 2
/ / \ \
2 1 2 3
96. Unique Binary Search Trees 给定一个自然数n,求由1-n组成的不同的二叉搜索树的个数。这道题可以用动态规划解决。设dp(i)表示输入为i时的解,则该问题的解为dp(n)。设f(i)表示以i为根的二叉搜索树的个数,则有:
dp(n) = f(1) + f(2) +f(3) + ... + f(n)
f(i) = dp(i-1) * dp(n-i)
整理上述式子,得
dp(n) = dp(0)*dp(n-1) + dp(1)*dp(n-2) + dp(2)*dp(n-3) + ... + dp(n-1)*dp(0)
代码实现如下:
class Solution {
public int numTrees(int n) {
if (n == 0) {
return 0;
}
int[] dp = new int[n + 1];
dp[0] = 1;
dp[1] = 1;
for (int i = 2; i <= n; i++) {
for (int j = 0; j < i; j++) {
dp[i] += dp[j] * dp[i - 1 - j];
}
}
return dp[n];
}
}