96. Unique Binary Search Trees.md

96. Unique Binary Search Trees

Question

Given n, how many structurally unique BST's (binary search trees) that store values 1 ... n?

Example:

Input: 3
Output: 5
Explanation:
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

Thinking:

• Method 1:DP 分析：
  1. 假设我们有[1,2,3,4,5,6,7,8], 我们可以选取其中任何一个为根节点。
  2. 当n = 0时，为1；
  3. 当n = 1时，为1；
  4. 当n = 2时，数组为[1,2],根节点可以为1,2。dp[2] = dp[0] * dp[1] + dp[1] * dp[0].
  5. n = 3, dp[3] = dp[0] * dp[2] + dp[1] * dp[2] + dp[2] * dp[0] 每一项分析：左边的可能性个数 * 右边的可能性个数

class Solution {
    public int numTrees(int n) {
        int[] dp = new int[n + 1];
        dp[0] = 1;
        dp[1] = 1;
        for(int i = 2; i <= n; i++){
            int sum = 0;
            for(int j = 0; j < i; j++)
                sum += dp[j] * dp[i - 1 - j];
            dp[i] = sum;
        }
        return dp[n];
    }
}

二刷

二刷的时候隐约能记得大致的思路，但是还是参考了一刷的结果。

class Solution {
    public int numTrees(int n) {
        int[] dp = new int[n + 1];
        dp[0] = 1; dp[1] = 1;
        for(int i = 2; i <= n; i++){
            int sum = 0;
            for(int j = 0; j < i; j++)
                sum += dp[j] * dp[i - 1 - j];
            dp[i] = sum;
        }
        return dp[n];
    }
}

Third Time

• Method 1: dp

   class Solution {
   	public int numTrees(int n) {
   		int[] dp = new int[n + 1];
   		dp[0] = 1; dp[1] = 1;
   		for(int i = 2; i <= n; i++){    //O(n ^ 2) 
   			// left + 1 + right
   			// range of left [0, i - 1], right = i - 1 - left
   			for(int left = 0; left <= i - 1; left++){   
   				dp[i] += dp[left] * dp[i - 1 - left];
   			}
   		}
   		return dp[n];
   	}
   }