Leetcode 96. Unique Binary Search Trees[medium]
2016-11-27 19:57
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题目:
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
区间dp。
因为二叉搜索树的性质:根的左孩子比根小,根的右孩子比根大。所以没有后效性,可以dp。
dp[i][j]表示使用i~j这(j - i + 1)个数组生成的不同的二叉搜索树。
dp[i][j] = sigma(dp[i][k - 1] * dp[k + 1][j])
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
区间dp。
因为二叉搜索树的性质:根的左孩子比根小,根的右孩子比根大。所以没有后效性,可以dp。
dp[i][j]表示使用i~j这(j - i + 1)个数组生成的不同的二叉搜索树。
dp[i][j] = sigma(dp[i][k - 1] * dp[k + 1][j])
class Solution { public: #define N 555 int dp ; int numTrees(int n) { memset(dp, -1, sizeof dp); dp[1] = dfs(1, n); return dp[1] ; } int dfs(int lt, int rt) { if (rt < lt) return 1; if (dp[lt][rt] != -1) return dp[lt][rt]; dp[lt][rt] = 0; for (int i = lt; i <= rt; i++) { dp[lt][i - 1] = dfs(lt, i - 1); dp[i + 1][rt] = dfs(i + 1, rt); dp[lt][rt] += dp[lt][i - 1] * dp[i + 1][rt]; } return dp[lt][rt]; } };
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