Garvin的专栏

背景（以下背景资料转载自：http://www.cnblogs.com/springfor/p/3874591.html?utm_source=tuicool）DFS（Dpeth-first Search）顾名思义，就是深度搜索，一条路走到黑，再选新的路。记得上Algorithm的时候，教授举得例子就是说，DFS很像好奇的小孩，你给这个小孩几个盒子套盒子，好奇的小孩肯定会一个盒子打开后继续再在这个

题目Given a binary tree, flatten it to a linked list in-place.For example,Given 1 / \ 2 5 / \ \ 3 4 6The flattened tree should look like: 1 \ 2

题目Given a binary tree, find its maximum depth.The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.代码/** * Definition for binary tree * p

题目       Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).For example:Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 /

题目1.same treeGiven two binary trees, write a function to check if they are equal or not.Two binary trees are considered equal if they are structurally identical and the nodes have the same value.2.sym

题目Given a sorted linked list, delete all duplicates such that each element appear only once.For example,Given 1->1->2, return 1->2.Given 1->1->2->3->3, return 1->2->3.代码 public static ListNode deleteD

题目Given two sorted integer arrays A and B, merge B into A as one sorted array.Note:You may assume that A has enough space (size that is greater or equal to m + n) to hold additional elements from B. T

题目Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string.If the last word does not exist, return 0.Note: A word is defined

题目Given two binary strings, return their sum (also a binary string).For example,a = "11"b = "1"Return "100".代码public class Solution { public String addBinary(String a, String b) { S

题目The count-and-say sequence is the sequence of integers beginning as follows:1, 11, 21, 1211, 111221, ...1 is read off as "one 1" or 11.11 is read off as "two 1s" or 21.21 is read off as "one 2, then

题目Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.For example:Given the below binary tree and sum =

1.时间复杂度O(N),内存O(1)的效率下实现单链表的翻转public static TreeNode revers(TreeNode head){ TreeNode temp,first,second; first=head; second=head.next; while(second!=null){ temp=second.next; second.next=

题目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) must be

题目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), d

题目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 \

题目Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.An example is the root-to-leaf path 1->2->3 which represents the number 123.Find the total sum of

题目Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also known as the Hamming weight).For example, the 32-bit integer ’11' has binary representation 0000000000

题目Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.You may assume that the array is non-empty and the majority element alwa

题目Rotate an array of n elements to the right by k steps.For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].Note:Try to come up as many solutions as you can, the

题目Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. Fo

题目Given numRows, generate the first numRows of Pascal's triangle.For example, given numRows = 5,Return[ [1], [1,1], [1,2,1], [1,3,3,1], [1,4,6,4,1]]代码public class Solution { public

题目Write a program to find the node at which the intersection of two singly linked lists begins.For example, the following two linked lists:A: a1 → a2 ↘

题目Implement strStr().Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.代码public class Solution { public int strStr(String haystack, String ne

题目    Given an array and a value, remove all instances of that value in place and return the new length.The order of elements can be changed. It doesn't matter what you leave beyond the new length.代码p

题目      这道题是链表的简单应用，将两个有序链表合成一个有序链表。       思路是：表一，表二各取两个对象，分别指向current和next，进行交叉比较排序。Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nod

1.理解            对于递归函数的理解，我觉得是比较重要的，因为很多大神能把递归函数用的惟妙惟肖，不光是他们的编程功力高深，更主要是能理解这个算法。比较直白的理解是，如果一个事件的逻辑可以表示成，f(x)=nf(x-1)+o(x)形式，那么就可以用递归的思路来实现。编写递归逻辑的时候要知道如下法则：1.要有基准  比如说，f(x)=f(x-1)+1，如果不加入基准，f(0)的值是多少，

1.背景             最大序列和问题一直以来是一个比较经典的算法题，看到这个问题，有很多解题的办法。今天看到了一种时间复杂度只为O(n)的解题算法，在这里记录下。             思路很简单，比方说有P1,P2,P3,P4.....这样一个序列，我们从P1开始求和，比如说在P5时求和数小于零，就可以断定。第一种情况，最大序列在P1~P5之间，或者说在P6~Pn之间。因为如果P1

1.背景           欧几里得算法是一个求最大因子的快速算法。如果m,n存在最大因子k，假设m=x*n+r,那么m和n可以整出k的话，r也肯定可以整除k           因为定理：如果M>N,则M mod N2.代码           package Algorithm_analysis;public class Euclid { public static void mai

（转载请注明出处：http://blog.youkuaiyun.com/buptgshengod）1.背景        装饰模式就是为一个对象增添新的功能，在对象的基础上修饰。本文实现的功能是，对象a有方法show输出“This is a test”，经过装饰后输出“This is a test for decorate”。实质就是多个类通用一个接口，然后将要修饰的类定义在修饰类的构造函数里。2

（转载请注明出处：http://blog.youkuaiyun.com/buptgshengod）1.题目介绍   中缀表达式是将运算符放在运算数中间的写法，如a+b*c。后缀表达式是将运算符放在运算数后面，如abc*+。2.代码实现部分 import java.util.Stack; public class Main { private String t

（转载请注明出处：http://blog.youkuaiyun.com/buptgshengod）1.题目介绍   二叉树是一种基本的数据结构。查找二叉树是一种方便与查找，删除，插入等功能的二叉树，它要求每个父节点的左分支小于父节点，右分支大于父节点。下面我们来实现下面这个查找二叉树。2.java代码实现public class BinaryTree { private N

（转载请注明出处：http://blog.youkuaiyun.com/buptgshengod）1.题目           这是一道检测inversion count的算法。它将检测输入序列中反序输入的个数,即检测其中有几对A[i] > A[j], i 比如输入4，3，2，1，输出应该为3+2+1=6.。 因为：1.  4比3，2，1大，但4在输入序列中是第一位，比3，2，1的inde

（转载请注明出处：http://blog.youkuaiyun.com/buptgshengod）1.题目     给定一个数字序列，其中有正有负，确定最大子序列和。用穷举法最好的结果也是时间复杂度O(n²)。后来看到一个聪明的方法，直接使时间复杂度变为O(n)。2.解法（1）穷举法       把所有序列都算出来找到最大的。/* 最大序列和问题的求解，一组数列有正有负，找出其中

（转载请注明出处：http://blog.youkuaiyun.com/buptgshengod）1.背景知识       大O标记就不用我说了吧。O(n)这种时间复杂度的意义自己google吧。这里简单讲下从代码推算。2.具体案例 (1)案例一  int a=0; //第一行for(int i=0;i<=N;i++)//第二行{ a+=i

1.背景       以一个题目为例，一个整数x是一组按大小顺序排列好的数列中的一个数，我们要找到x在数列中的索引位置。比如按从小到大排列的数列：-3,-2,0,4,5,7,12,64我们要找到数字7的位置，如果是线性查找，时间复杂度是O(n),如果用折半查找的话，时间复杂度是O(log(n))，因为每次折半，计算量少一半，所以取对数。2.代码package Algorithm_analysis;

1.背景          单链表是最基本的数据结构，仔细看了很久终于搞明白了，差不每个部分，每个链都是node的一个对象。需要两个参数定位：一个是index，表示对象的方位。另一个是node的对象。2.代码node类public class Node { protected Node next; protected int data; public Node(int data){

1.背景     追随着buptwusuopu大神的脚步，最近在研习动态规划。动态规划应该叫一种解决问题的思想，记得又一次去某公司面试就被问到了这个。       多于动态规划的理解，大致是这样的，从空集合开始，每增加一个元素就求它的最优解，直到所有元素加进来，就得到了总的最优解。           比较典型的应用就是背包问题，有一个重量一定的包，有若干件物品，他们各自有不同的重量和价值，怎样背

题目     题目要求：去除sort int数组中的重复项。     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,

题目        这是道链表的简单应用题目，删除从结尾数第n个节点。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

题目          爬楼梯问题，这是一道很有趣的问题。首先看题目：You are climbing a stair case. It takes n steps to reach to the top.Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?这