Generate Parentheses (Java)

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example, given n = 3, a solution set is:

"((()))", "(()())", "(())()", "()(())", "()()()"




Solution: Greedy and Recursive

Take advantage of the feature of parentheses, left side and right side must match each other,  in other words, if there a left side parenthese,  whatever position it is, there must be a right side parenthese to match it.

Best Time to Buy and Sell Stock (Java)

Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Solution:
loop through entire list  and  catch min_price and the position of this this price, when current reached price's position after min_price's position then do calculate difference and up max_profit

Jump Game (Java)

Leetcode

Given an array of non-negative integers, you are initially positioned at the first index of the array.

Each element in the array represents your maximum jump length at that position.

Determine if you are able to reach the last index.

For example:
A = [2,3,1,1,4], return true.

A = [3,2,1,0,4], return false.

Jump Game II (java)

LeetCode

Given an array of non-negative integers, you are initially positioned at the first index of the array.

Each element in the array represents your maximum jump length at that position.

Your goal is to reach the last index in the minimum number of jumps.

For example:
Given array A = [2,3,1,1,4]

The minimum number of jumps to reach the last index is 2. (Jump 1 step from index 0 to 1, then 3 steps to the last index.)

Solution: Greedy

At first, I try to solve this problem with DFS, but exceeded the time limitation, then I search the Internet find a very good solution for this question - Greedy Algorithm. the main idea is try to find the longest distance by each jump can reach and check if this distance can pass the total length of this array, of course we should have a variable to keep record of the current steps. if this distance cannot pass the total length of this array, then we should go through all the position within this distance to see if it can pass the array by jumping from there

I think code should be more clear than my description 

Gas Station (Java)

Leetcode