96. Unique Binary Search Trees LeetCode
2016-02-13 18:24
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题意：求n个节点能生成不同搜索二叉树的方案数。
题解：dp,dp[i] 表示有i个节点的时候能生成不同的搜索二叉树的方案数。那么转移等于dp[i] = dp[j - 1] * dp[i - j] (1<= j <= i).也就是左边j - 1个节点，右边 i - j个节点。
题解：dp,dp[i] 表示有i个节点的时候能生成不同的搜索二叉树的方案数。那么转移等于dp[i] = dp[j - 1] * dp[i - j] (1<= j <= i).也就是左边j - 1个节点，右边 i - j个节点。
class Solution {
public:
int numTrees(int n) {
int dp[n + 1];
memset(dp,0,sizeof(dp));
dp[0] = dp[1] = 1;
dp[2] = 2;
for(int i = 3; i <= n; i++)
{
for(int j = 1; j <= i; j++)
{
dp[i] += dp[j - 1] * dp[i - j];
}
}
return dp
;
}
};
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