Unique Binary Search Trees (6行代码 dp)
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大神的解释:DP solution in 6 lines with explanation"
歌词大意:
Given n, how many structurally unique BST's (binary search trees) that store values 1...n?
For example,
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
\ / / / \ \
3 2 1 1 3 2
/ / \ \
2 1 2 3
G(n)
: the number of unique BST for a sequence of length nG(n) = F(1, n) + F(2, n) + ... + F(n, n)
空树:G(0)=1
只有一个根节点:G(1)=1
F(i, n), 1 <= i <= n
: 根是i的二分搜索树的数量,i为根,左子树有i-1个节点,右子树有n-i个节点,因此,i为根的情况下,有G(i-1)*G(n-i)中情况G(n) = F(1, n) + F(2, n) + ... + F(n, n)
= G(0) * G(n-1) + G(1) * G(n-2) + … + G(n-1) * G(0)
public int numTrees(int n) {
int [] G = new int[n+1];
G[0] = G[1] = 1;
for(int i=2; i<=n; ++i) {
for(int j=1; j<=i; ++j) {
G[i] += G[j-1] * G[i-j];
}
}
return G[n];
}
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