longest increasing continuous subsequence in a 2D matrix

craiglin1992 — Thu, 17 Sep 2015 02:16:48 +0800

dp + dfs //These two are only for traversing around x,y public int[] xMove = { 1, 0, 0, -1 }; public int[] yMove = { 0, 1, -1, 0 }; public int longestIncreasingContinuousSubsequenceII(int

perfect squares find the least number of perfect square numbers (1, 4, 9, 16, ...) which sum to n

craiglin1992 — Thu, 17 Sep 2015 02:01:21 +0800

用动态规划Dynamic Programming来做，我们建立一个长度为n+1的一维dp数组，将第一个值初始化为0，其余值都初始化为INT_MAX, i从0循环到n，j从1循环到i+j*j <= n的位置，然后每次更新dp[i+j*j]的值，动态更新dp数组，其中dp[i]表示正整数i能少能由多个完全平方数组成，那么我们求n，就是返回dp[n]即可，也就是dp数组的最后一个数字，参见代码如下：int

missing ranges of sorted integer array

craiglin1992 — Thu, 17 Sep 2015 00:53:29 +0800

大自然的搬运工：http://www.danielbit.com/blog/puzzle/leetcode/leetcode-missing-ranges Given a sorted integer array where the range of elements are [0, 99] inclusive, return its missing ranges. For example, g

find maximized count of 0 on left and 1 on right in binary array

craiglin1992 — Thu, 17 Sep 2015 00:48:19 +0800

给定一个 array,只包含 0,1, 找到一个分割位置,使左侧 0 出现的个数和右侧 1 出现的个数之和最大化brute force, time: O(n^2), space: O(1) follow up, can we solve in O(n): DP, two temp array-google 1point3acres follow up, can we use O(1) spac

serialize a list of string and deserialize it

craiglin1992 — Thu, 17 Sep 2015 00:46:41 +0800

import java.util.*; public class combineStrings{ public static void main(String[] args) { String[] arr = {"abc%cde", "a#aa", "haha"}; for(String s : arr) {

wiggle sort Given a list of integers, sort them so the output is s1 <= s2 >=s3 <=s4 ... sN.

craiglin1992 — Thu, 17 Sep 2015 00:18:54 +0800

每个元素和后一个元素比较看是不是希望的次序，如果不是的话互换就好public void wiggle_sort(int[] arr) { int n = arr.length; if(n <= 1) return; boolean inc = true; int prev = arr[0]; for(int i=1; i

Reservoir Sampling (a random element from indefinite length stream with same probability)

craiglin1992 — Thu, 17 Sep 2015 00:07:01 +0800

Reservoir Sampling Reservoir sampling is a family of randomized algorithms for randomly choosing k samples from a list of n items, where n is either a very large or unknown number. Typically n is larg

calculate sorted y of function y = ax^2 + bx + c given sorted x array

craiglin1992 — Wed, 16 Sep 2015 15:42:31 +0800

给定一个一元二次方程式y = ax^2 + bx + c, 一个sorted array X, 将X中所有元素代入方程中，返回sorted的Ysolution: 看a< or > 0,，搞出两边的y们，然后mergepublic class Ax2 { public static void main(String[] args) { } // x is sorted array

Randow Id generator with array of probability

craiglin1992 — Wed, 16 Sep 2015 15:29:42 +0800

给一组id和表示每个id出现概率的数组，概率之和为1.要求随机生成id，使得随机出的id满足之前的概率数组。**如果id很多，调用这个随机生成方法的次数也很多，怎么优化使用prefix sum，使用binary searchclass RandomID { public static void main(String[] args) { int[] id = { 1, 2,

different/Unique/distinct Binary Search Trees

craiglin1992 — Wed, 16 Sep 2015 15:24:05 +0800

Given n, how many structurally unique BST's (binary search trees) that store values 1...n?For example, Given n = 3, there are a total of 5 unique BST's. 1 3 3 2 1 \ /

adding up two (sum) big/large number with string representation

craiglin1992 — Wed, 16 Sep 2015 11:30:41 +0800

public static String sumStrings(String a, String b) { char[] num1 = a.toCharArray(); char[] num2 = b.toCharArray(); int i = num1.length - 1; int j = num2.length - 1;

Find a sorted subsequence of size 3 three in linear time

craiglin1992 — Wed, 16 Sep 2015 10:53:29 +0800

Given an array of n integers, find the 3 elements such that a[i] < a[j] < a[k] and i < j < k in 0(n) time. If there are multiple such triplets, then print any one of them.Examples:Input: arr[] = {12,

Data Structure for HTML DOM with compare text inside

craiglin1992 — Wed, 16 Sep 2015 10:20:35 +0800

1. 设计HTML DOM数据结构信息是

Hello

或者

Hello

【提示是类似tree】 2. 写一个方法比较两个HTML数据中存粹文本是否一样. 鍥磋鎴戜滑@1point 3 acres 上面两个例子都是Hello，所以此方法返回truepub

compute exact number of triples of distinct elements in large array

craiglin1992 — Wed, 16 Sep 2015 10:10:30 +0800

Given an array of N=10^6 int32 and an int32 X, compute the exact number of triples (a, b, c) of distinct elements of the array so that a + b + c <= X public int threeSumSmaller(int[] nums, int targe

fast Update, sum operation of 2D matrix

craiglin1992 — Wed, 16 Sep 2015 09:45:27 +0800

Given a 2D space of maximum size NxN which supports two operations : [1] void UPDATE(x,y,v) - sets the value of cell [x,y] to v [2] int SUM(x1,y1,x2,y2) - returns sub-rectangle sum (x1,y1) to (x2

Find the longest substring with k unique distinct characters in a given string (Google)

craiglin1992 — Wed, 16 Sep 2015 09:03:14 +0800

The problem can be solved in O(n). Idea is to maintain a window and add elements to the window till it contains less or equal k, update our result if required while doing so. If unique elements exceeds

distinct sequence Google

craiglin1992 — Wed, 16 Sep 2015 08:35:14 +0800

leetcode ,暴力解法dfs, 双指针，不行 public int numDistinct(String s, String t) { return numDistinctHelper(s,t,0,0); } public int numDistinctHelper(String s, String t, int si, int ti){

Rearrange int array to put odd number to left and even to right

craiglin1992 — Wed, 16 Sep 2015 03:30:07 +0800

双指针，左右调换。no extra memory, O(n)class Solution: def sepOddEven(self, nums): if not nums or len(nums) == 0: return None left = 0 right = len(nums) - 1 while left < righ

float number square root

craiglin1992 — Wed, 16 Sep 2015 02:10:13 +0800

public float sqrt(float f, int p)， precision是表示小数点后位数（2就要两位一致） p is the number of digits after the decimal point has to be rightSolution: 判断数据是否符合精确度，就用两个数都乘以10^p，再取整比较是否相等。比如p=2， f=0.64, curRes =

word abbreviation

craiglin1992 — Tue, 15 Sep 2015 23:41:41 +0800

一个string只保留首尾字母, 中间字母数用数字表示, 比如 “abcde” = “a3e”, “ab” = “ab”, “abc” = “a1c” . 给一组string, 输出相同abbreviation的所有string, 但重复的string不算. abv_string = StringBuilder.append(charAt[0]), StringBuilder.append(str.

Pushing Code