guolinaileen’s gists

guolinaileen / 3Sum Closest.cpp

Last active December 11, 2015 19:08

	class Solution {
	public:
	int threeSumClosest(vector<int> &num, int target) {
	// Start typing your C/C++ solution below
	// DO NOT write int main() function
	sort(num.begin(), num.end());
	int result=num[0]+num[1]+num.back();
	for(int i=0; i<num.size(); i++)
	{
	int j=i+1, k=num.size()-1;

guolinaileen / Letter Combinations of a Phone Number.java

Last active December 11, 2015 19:48

recursively solve this problem; uninitialized char equals to 0

	public class Solution {
	public static ArrayList<String> letterCombinations(String digits) {
	if(digits==null) return null;
	ArrayList<String> result=new ArrayList<String>();
	if(digits.length()==0)
	{
	result.add("");
	return result;
	}
	char [][] maps=new char[10][4];

guolinaileen / Remove Nth Node From End of List.java

Last active December 11, 2015 19:48

Try to do this in one pass.

-/**
- * Definition for singly-linked list.
- * public class ListNode {
- *     int val;
- *     ListNode next;
- *     ListNode(int x) {
- *         val = x;
- *         next = null;
- *     }
- * }

guolinaileen / gist:4651024

Created January 27, 2013 22:34

Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid. The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.

	public class Solution {
	public boolean isValid(String s) {
	int length=s.length();
	char [] array=s.toCharArray();
	if(length==0) return true;
	Stack<Character> stack=new Stack<Character>();
	for(int i=0; i<length; i++)
	{
	if(array[i]=='(' \|\| array[i]=='{' \|\| array[i]=='[' )
	{

guolinaileen / Generate Parentheses.java

Last active December 11, 2015 20:09

Generate ParenthesesFeb 13 '12 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: "((()))", "(()())", "(())()", "()(())", "()()()"

	public class Solution {
	public ArrayList<String> generateParenthesis(int n) {
	ArrayList<String> result=new ArrayList<String>();
	char [] array=new char[2*n];
	generate(result, n, n, array, 0);
	return result;
	}
	void generate(ArrayList<String> result, int left, int right, char [] array, int num)
	{
	if(left==0 && right==0 )

guolinaileen / Merge k Sorted Lists.java

Created January 28, 2013 08:18

This method should be changed. PriorityQueue can be used to solve this problem.

	/**
	* Definition for singly-linked list.
	* public class ListNode {
	* int val;
	* ListNode next;
	* ListNode(int x) {
	* val = x;
	* next = null;
	* }
	* }

guolinaileen / Swap Nodes in Pairs.java

Last active December 26, 2019 08:24

Given a linked list, swap every two adjacent nodes and return its head. For example, Given 1->2->3->4, you should return the list as 2->1->4->3. Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

	/**
	* Definition for singly-linked list.
	* public class ListNode {
	* int val;
	* ListNode next;
	* ListNode(int x) {
	* val = x;
	* next = null;
	* }
	* }

guolinaileen / Swap Nodes in Pairs.cpp

Last active December 11, 2015 20:18

	/**
	* Definition for singly-linked list.
	* struct ListNode {
	* int val;
	* ListNode *next;
	* ListNode(int x) : val(x), next(NULL) {}
	* };
	*/
	class Solution {
	public:

guolinaileen / Best Time to Buy and Sell Stock II.java

Last active December 11, 2015 21:09

	public class Solution {
	public int maxProfit(int[] prices) {
	// Start typing your Java solution below
	// DO NOT write main() function
	if(prices.length==0) return 0;
	int profit=0;
	for(int i=1; i<prices.length; i++)
	{
	if(prices[i]>prices[i-1])
	{

guolinaileen / Best Time to Buy and Sell Stock.java

Last active December 11, 2015 21:09

	public class Solution {
	public int maxProfit(int[] prices) {
	// Start typing your Java solution below
	// DO NOT write main() function
	if(prices.length==0) return 0;
	int min=prices[0], max=prices[0];
	int profit=0;
	for(int i=1; i<prices.length; i++)
	{
	if(prices[i]>max)