Wu_George的专栏

概念：  编译型语言：把做好的源程序全部编译成二进制代码的可运行程序。然后，可直接运行这个程序。  解释型语言：把做好的源程序翻译一句，然后执行一句，直至结束！  区别：  编译型语言，执行速度快、效率高；依赖编译器、跨平台性差些。如C、C++、Delphi、Pascal，Fortran

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.Each number in C may only be used once in the combina

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

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.The same repeated number may be chosen from C unlimited numb

转自：http://www.cnblogs.com/reynold-lei/p/3370026.html字符串匹配的KMP算法举例来说，有一个字符串"BBC ABCDAB ABCDABCDABDE"，我想知道，里面是否包含另一个字符串"ABCDABD"？ 许多算法可以完成这个任务，Knuth-Morris-Pratt算法（简称KMP）是最常用的之一。它以三个发明者命名，

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].解法一：DFS，递归时当前元素

Given a string s, partition s such that every substring of the partition is a palindrome.Return all possible palindrome partitioning of s.For example, given s = "aab",Return [ ["aa","

思路参考：http://www.oschina.net/code/snippet_112092_9454

Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and set.get(key) - Get the value (will always be positive) of the key if

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 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 quadruplets in the array which gives the sum of target.Note:Element

Divide two integers without using multiplication, division and mod operator.题意：不适用*public class Solution { public int divide(int a, int b) { if(a == 0 || b==0) return 0; if(b

Suppose a sorted array is rotated at some pivot unknown to you beforehand.(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).You are given a target value to search. If found in the array retur

描述You are given a sequence of integers, A = a1, a2, ... an. A consecutive subsequence of A (say ai, ai+1 ... aj) is called a "repeated sequence" if it appears more than once in A (there exists

Given a collection of numbers that might contain duplicates, return all possible unique permutations.For example,[1,1,2] have the following unique permutations:[1,1,2], [1,2,1], and [2,1,1].

MS的一个面试题，删除链表中连续重复的节点，如1,2,2,2,3

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 collection of integers that might contain duplicates, S, return all possible subsets.Note:Elements in a subset must be in non-descending order.The solution set must not contain dupli

全手写char *strcpy(char* strDest, const char* strSrc)char *strcpy(char *strDest, const char *strSrc){ if(strDest==NULL || strSrc==NULL) return NULL; if(strDest == strSrc) return strDest; char *dest

int main(){ char *s = "abc"; char d[3]; strcpy(d,s); printf("%s",d);//注意，可以正常输出abc，然后由于\0的关系发生了堆栈溢出，覆盖了返回地址，导致运行时错误，程序崩溃 return 0;}

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).For example, this binary tree is symmetric: 1 / \ 2 2 / \ / \3 4 4 3But the f

转自：http://blog.youkuaiyun.com/luckyxiaoqiang/article/details/8937978二分查找，最基本的算法之一，也是面试中常被考察的重点，因为基本的算法最能反映出一个人的基础是否扎实。本文对二分查找相关题目做一个总结。题目列表：1. 给定一个有序（非降序）数组A，求任意一个i使得A[i]等于target，不存在则返回-12.

转自：http://blog.youkuaiyun.com/walkinginthewind/article/details/7393134链表是最基本的数据结构，面试官也常常用链表来考察面试者的基本能力，而且链表相关的操作相对而言比较简单，也适合考察写代码的能力。链表的操作也离不开指针，指针又很容易导致出错。综合多方面的原因，链表题目在面试中占据着很重要的地位。本文对链表相关的面试题做了较为全面

二叉树的遍历-递归与非递归         二叉树是一种非常重要的数据结构，很多其它数据结构都是基于二叉树的基础演变而来的。对于二叉树，有前序、中序以及后序三种遍历方法。因为树的定义本身就是递归定义，因此采用递归的方法去实现树的三种遍历不仅容易理解而且代码很简洁。而对于树的遍历若采用非递归的方法，就要采用栈去模拟实现。在三种遍历中，前序和中序遍历的非递归算法都很容易实现，非递归后序遍历实现

题意：二叉树的前序遍历这里先提供递归实现，后续补上public ArrayList preorderTraversal(TreeNode root) { if(root == null) return new ArrayList(); ArrayList list = new ArrayList(); ArrayList l

Given a string containing only digits, restore it by returning all possible valid IP address combinations.For example:Given "25525511135",return ["255.255.11.135", "255.255.111.35"]. (Order

转自：http://blog.youkuaiyun.com/walkinginthewind/article/details/7518888树是一种比较重要的数据结构，尤其是二叉树。二叉树是一种特殊的树，在二叉树中每个节点最多有两个子节点，一般称为左子节点和右子节点（或左孩子和右孩子），并且二叉树的子树有左右之分，其次序不能任意颠倒。二叉树是递归定义的，因此，与二叉树有关的题目基本都可以用递归思想

题意：二叉树的后续遍历这里先使用递归的形式实现，以后再数据结构中补上非递归实现，对边界条件的处理要想好！public ArrayList postorderTraversal(TreeNode root) { if(root == null) return new ArrayList(); ArrayList list = new ArrayList();

一、递归实现：public ArrayList inorderTraversal(TreeNode root) { ArrayList list = new ArrayList(); if(root == null) return list; ArrayList leftList = inorderTraversal(root.

Given n, generate all structurally unique BST's (binary search trees) that store values 1...n.For example,Given n = 3, your program should return all 5 unique BST's shown below. 1 3

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

Given a set of distinct integers, S, return all possible subsets.Note:Elements in a subset must be in non-descending order.The solution set must not contain duplicate subsets.For exa

Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.For example,If n = 4 and k = 2, a solution is:[ [2,4], [3,4], [2,3], [1,2], [1,3], [1,4],]

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?public class Solution { public void rotate(int[