leetcode Guess Number Higher or Lower II
We are playing the Guess Game. The game is as follows:
I pick a number from 1 to n. You have to guess which number I picked.
Every time you guess wrong, I’ll tell you whether the number I picked is higher or lower.
However, when you guess a particular number x, and you guess wrong, you pay $x. You win the game when you guess the number I picked.
Example:
123456789 n = 10, I pick 8.First round: You guess 5, I tell you that it's higher. You pay $5.Second round: You guess 7, I tell you that it's higher. You pay $7.Third round: You guess 9, I tell you that it's lower. You pay $9.Game over. 8 is the number I picked.You end up paying $5 + $7 + $9 = $21.Given a particular n ≥ 1, find out how much money you need to have to guarantee a win.
Hint:
 The best strategy to play the game is to minimize the maximum loss you could possibly face. Another strategy is to minimize the expected loss. Here, we are interested in the first scenario.
 Take a small example (n = 3). What do you end up paying in the worst case?
 Check out this article if you’re still stuck.
 The purely recursive implementation of minimax would be worthless for even a small n. You MUST use dynamic programming.
 As a followup, how would you modify your code to solve the problem of minimizing the expected loss, instead of the worstcase loss?
dynamic programming
leetcode Coin Change
leetcode Coin Change
You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return
1
.Example 1:
coins =[1, 2, 5]
, amount =11
return3
(11 = 5 + 5 + 1)Example 2:
coins =[2]
, amount =3
return1
.Note:
You may assume that you have an infinite number of each kind of coin.
leetcode Best Time to Buy and Sell Stock with Cooldown
leetcode Best Time to Buy and Sell Stock with Cooldown
Say you have an array for which the i^{th} element is the price of a given stock on day i.
Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times) with the following restrictions:
 You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
 After you sell your stock, you cannot buy stock on next day. (ie, cooldown 1 day)
Example:
prices = [1, 2, 3, 0, 2]
maxProfit = 3
transactions = [buy, sell, cooldown, buy, sell]
leetcode 动态规划（DP）
本次题解包括

 53 Maximum Subarray
 62 Unique Paths
 63 Unique Paths II
 64 Minimum Path Sum
 70 Climbing Stairs
 72 Edit Distance
 91 Decode Ways
 97 Interleaving String
 115 Distinct Subsequences
 139 Word Break
 140 Word Break II
 152 Maximum Product Subarray
 174 Dungeon Game
 198 House Robber
 213 House Robber II
 221 Maximal Square
 712. Minimum ASCII Delete Sum for Two Strings
 718. Maximum Length of Repeated Subarray
leetcode Best Time to Buy and Sell Stock
本次题解包括
 121 Best Time to Buy and Sell Stock
 122 Best Time to Buy and Sell Stock II
 123 Best Time to Buy and Sell Stock III
 188 Best Time to Buy and Sell Stock IV
 714. Best Time to Buy and Sell Stock with Transaction Fee