LeetCode之Unique Binary Search Trees
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/*动态规划。参考自:https://github.com/soulmachine/leetcode设dp[n]表示n个没有重复的数组组成的二叉搜索数的数目。易知,以某个数字i为根节点的二叉树数目为其左子树数目乘以右子树数目。所以有dp[n] = sum(dp[i]*dp[n-1-i]),其中(i = 0,1,2...n-1)。*/class Solution {public: int numTrees(int n) { vector<int> dp(n+1, 0); dp[0] = dp[1] = 1; for(int i = 2; i < n+1; ++i){ for(int j = 0; j < i; ++j){ dp[i] += dp[j]*dp[i-1-j]; } } return dp[n]; }};
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