Leetcode 题解 - Dynamic Programming 1D

题解

State: (n)

State(n): how many unique BST to form with n nodes from 1, n, where i is the selected root

for range 1, i -1, this is calculated range for unique BST

for rangei, n-i, this is the another range for unique BST

multiply the answer for above two ranges for final result

High Level:

1. Initialize memo with n + 1 slots
2. call dfs(n)

dfs(n):

1. Base case (n <= 1) -> return 1
2. If memon != null -> return memon
3. Ask subproblems for answers:

a. for i in 1,n:

i. left = dfs(i-1), right = dfs(n-i)

ii. result = left * right

4. Update memon and return result

时间复杂度：O(n**2)

另可以用正向填表：把dfs的flow反过来，从底层子问题开始往大计算，最终计算到顶层问题

High Level

1. Initialize memon + 1
2. Fill out base case -> memo0 = memo1 = 1
3. for i in 2,n:

a. Use Transition Rule -> for j in 1, i:

i. memoi += memoj-1 * memoi - j

4. return memon

时间复杂度：O(n**2)

思路

High Level

1. Initialize memo

2. Call dfs(s,n)

dfs(s,n)

1. Base case -> n <= 1 return 1
2. If memon != null -> return memon
3. Ask Subproblems for answers

a. if x is valid (not 0) -> result += memon-1

b. if xx is valid (<= 26) -> result += memon-2

4. Update memo and return result