ass673556617的专栏

先将快速排序的代码放上：public class QuickSort {public void quickSort( int[] List, int a, int b){if(aint position=partition(List,a,b);quickSort(List,a,position-1);quickSort(List,positi

先附上代码：public class Heapify {//从n/2开始查找public void buildHeapify(int[] list){//从一开始存//int[] list ={0,4,2,6,24,12,9};  length=7  max=6int start=(list.length-1)/2;for

采用子底向上的方式，依次计算切割钢管所能获得的最大收益，然后是对应的分割方案。

Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.Do not allocate extra space for another array, you must do this in place with

参考：http://blog.youkuaiyun.com/linhuanmars/article/details/20679963The count-and-say sequence is the sequence of integers beginning as follows:1, 11, 21, 1211, 111221, ...1 is read off as "one

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. Y

参考了网上的一种解题思路,原网址找不到了。。   思路的话仔细看看会明白的，当时举了一个例子按照这个流程跑了下。这个思路还是很清晰的。代码：package codes;public class MultiplyStrings { public static void main(String[] args) { String str1 = "9"; String str2 =

Determine whether an integer is a palindrome. Do this without extra space.实现代码：public boolean isPalindrome02(int x) { String num= Integer.toString(x); if(num.length()>11 || num.charAt(0)=='

Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Fin

Implement atoi to convert a string to an integer.Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input ca

提述：Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.写了一个算法，但是

实现思路参考了网上的内容。一开始的想法是通过一个嵌套循环计算每一位的最长子串，时间复杂度是O(n3)。在嵌套的第二个循环内，使用 map.containKey来判断当前是否存在值，存在的话跳出循环，不存在的话存入key。这个思路最大的问题是复杂度比较高。网上提供了 一种线性时间的解决方法，核心思想是当遇到重复的字符时，起点更新为重复字符的上一个位置。需要借助一个数组来保存上一个位置的信息。

Reverse digits of an integer.Example1: x = 123, return 321Example2: x = -123, return -321这道题其实不难，重要的是对一些特殊值进行判断。如果输入数字为超过int范围的值该如何处理。例如：1534236469，反过来会超范围，但是在计算的过程中不会提示你已经越界了。我用的是一个Long

The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)P A H NA P L S

leetecode 上的一道题：There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).实现代码：

You are given an n x n 2D matrix representing an image.Rotate the image by 90 degrees (clockwise).Follow up:Could you do this in-place?用个临时的数组，存下转换的位置再复制回去就行。代码：package codes;p

实现的时候参考了http://blog.sina.com.cn/s/blog_71d59f9a01018dh3.html 的思路。Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadru

Write a function to find the longest common prefix string amongst an array of strings.总之不难，但是脑袋要清醒..代码:public class Solution { public String longestCommonPrefix(String[] strs) {

Given an array of integers, find two numbers such that they add up to a specific target number.The function twoSum should return indices of the two numbers such that they add up to the target, whe

Given an integer, convert it to a roman numeral.Input is guaranteed to be within the range from 1 to 3999.这道题直接上来硬写的，感觉有点没动脑筋。想起当时用java课上一道关于扑克牌的课后作业，应该写的灵活些。像本题，如果还需要写出后续的数字这种写法就不妥当了。我想

Given a collection of numbers, return all possible permutations.For example,[1,2,3] have the following permutations:[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].实现的思路是使用递

Given a roman numeral, convert it to an integer.Input is guaranteed to be within the range from 1 to 3999.搞了挺久一直没搞出来。看了网上的参考后知道是推导的公式写错了。如果是从高位往地位扫（也就是一次循环用s.charAt(i)由高往低），那么相邻元素之

原网址：http://blog.youkuaiyun.com/linhuanmars/article/details/21354751发现递归跟动规的知识都还给老师了。。最近要大补一下。Given an array of non-negative integers, you are initially positioned at the first index of the a

参考：http://blog.youkuaiyun.com/linhuanmars/article/details/20092829参考了这上面的二分解法。自己写了一个最简单的发现会超时报错。关于递归方面一直不是很好，这学期要加强下算法这方面的知识。 关于代码中的几种情况用三个例子进行说明。代码：package codes;public class Pow {

Given a linked list, remove the nth node from the end of list and return its head.For example, Given linked list: 1->2->3->4->5, and n = 2. After removing the second node from the end, the

Given a digit string, return all possible letter combinations that the number could represent.A mapping of digit to letters (just like on the telephone buttons) is given below.Input:Digit st

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 va

最近补下动规跟递归。介绍动规的一个思路是先分析之前的一些通过递归解决的一些问题，判断他们在计算的过程中是否出现了过多的子问题的覆盖，如果覆盖了很多相同的结果那么计算势必要浪费很多时间在做无用功。动规采用自底向上的思路求解，一般在O(n2),O(n3)的复杂度内能够给出答案。实现的思路是先给出递归的解法，然后给出动规的解法。今天先以fibonacci 为例：代码：package

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.改了好久，可是明明不难的啊。代码：package leetcode;import java.uti

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.Note:Elements in a triplet (a,b,c

Given a sorted array of integers, find the starting and ending position of a given target value.Your algorithm's runtime complexity must be in the order of O(log n).If the target is not found

沿着结果网斜上方打印然后逆序就可以。代码：package codes;public class LongestCommonSubMemo { public static void main(String[] args) { String res = new LongestCommonSubMemo().lonestCommonSub("tofooodie", "

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.You may assume no duplicates in the array.

实现了一个动规的最长公共字串，晚上补上带纪录最长字串记录的算法。另外再补上一个递归的算法。代码：public class LongestCommonSub { public static void main(String[] args) { int [][]res = new LongestCommonSub().lonestCommonSub("tofooodie

有生之年希望快速排序不要再写错了。。注意一点，左右指针在往中间靠拢的时候，要先处理上边界。如果先处理下边界，那样的话是行不通的。举个例子就清楚些了。拿数组5，4，3，7，6，8 为例。例举了两个partition的过程。代码：package codes;public class MyQuickSort { public static void main(

Find the total area covered by two rectilinear rectangles in a 2D plane.Each rectangle is defined by its bottom left corner and top right corner as shown in the figure.Assume that the tota

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

最近在做个人博客，打算尝试下mongodb做数据库。记录下如何开启数据库的。1. 首先开启一个控制台，进入mongodb的安装目录的bin文件夹下，如图2.命令mongod --dbpath=数据库路径    路径自己选定就可以3.开启后会提示等待请求接入。可以用localhost:27017来查看状态这样就连上了。因为之前用的是My

问题描述可以见：http://datagenetics.com/blog/july22012/index.htmlYou are given two eggs, and access to a 100-storey building. Both eggs are identical. The aim is to find out the highest floo

看了下思路，实现了下。发现在将数组的两段合并的时候出了点问题，改了有十来分钟。之前在leetcode上做过一道题，将两个已经排序好的数组进行合并。归并排序的思路也是如此。归并排序采用的是递归的方法，在递归的最底层是一个个的单独的数字，两个数字按照顺序先合并，然后在进行过一次合并的这一层，再两两合并（此时要合并的两段已经是排序的了，因此用leetcode上的那道题的解题思路去实现就行）。由于合并