[LeetCode] 121: Unique Binary Search Trees
2017-09-10 20:53
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[Problem]
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
[Solution]
方法一
class Solution {
public:
int numTrees(int n) {
// Note: The Solution object is instantiated only once and is reused by each test case.
// invalid
if(n <= 0)return 0;
// initial
int dp[n+1];
memset(dp, 0, sizeof(dp));
dp[1] = 1;
for(int i = 2; i <= n; ++i){
dp[i] = 2*(2*(i-1)+1)*dp[i-1]/(i-1+2);
}
return dp
;
}
};
说明:版权所有,转载请注明出处。Coder007的博客
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
[Solution]
方法一
class Solution { public: int numTrees(int n) { // Note: The Solution object is instantiated only once and is reused by each test case. // invalid if(n <= 0)return 0; // initial int dp[n+1]; memset(dp, 0, sizeof(dp)); dp[0] = 1; for(int i = 1; i <= n; ++i){ for(int j = 1; j <= i; ++j){ dp[i] += dp[j-1]*dp[i-j]; } } return dp ; } };方法二:Catalan数 C_n+1= C_n * 2(2n+1)/(n+2)
class Solution {
public:
int numTrees(int n) {
// Note: The Solution object is instantiated only once and is reused by each test case.
// invalid
if(n <= 0)return 0;
// initial
int dp[n+1];
memset(dp, 0, sizeof(dp));
dp[1] = 1;
for(int i = 2; i <= n; ++i){
dp[i] = 2*(2*(i-1)+1)*dp[i-1]/(i-1+2);
}
return dp
;
}
};
说明:版权所有,转载请注明出处。Coder007的博客
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