Coding for Dreams

<br />#include <iostream>using namespace std;char qipan[8][8];bool col[8];int m,n,count;//r为行数，k代表已经填充的棋子数，count代表方案数void dfs(int r,int k){ if (k == m) { count++; return; } if (r > n) { return; } for (int i = 0;i < n;i

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 does not m

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].思路分析：这题是Permutations II问题的简

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 combination.N

Given a 2D board and a word, find if the word exists in the grid.The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically nei

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 number of

A knight moves on a chessboard two squares up, down, left, or right followed by one square in one of the two directions perpendicular to the first part of the move (i.e., the move is L-shaped). Suppos

Given a binary tree, find its minimum depth.The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.思路分析:这题和Maximum Depth of Binary Tree类似，但是现

Problem StatementYou are given a matrix with m rows and n columns of cells, each of which contains either 1or 0. Two cells are said to be connected if they are adjacent to each other horizontally, ver

Given 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.思路分析：判断两个树是否相同，基本也

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.思路分析：这题比较简单，关于树的题目通常都可以用递归解决，这题也不例外，递归解法的思

Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'.A region is captured by flipping all 'O's into 'X's in that surrounded region.For example,X X X XX O O XX X O XX O X X

Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.OJ's undirected graph serialization:Nodes are labeled uniquely.We use # as a separator for each node, and

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.For example:Given the following binary tree, 1

这是最近比较火的一个题目，因为一条推特的转播“Google HR：我们90%的工程师都用你写的软件，但是你竟然不会在白板上面反转一颗二叉树，所以滚吧”，各方看法不一，但看这个题目，的确是一个很简单的题目。考察最基本的递归和树操作。下面给出了递归实现和借助栈的迭代实现（非递归实现）。后一个版本的可扩展性更好，可以处理更大的树。实在是容易题，就是DFS遍历一遍树节点，把每个树节点的左右孩子互换就可以了。或许是那个ios开发牛人不屑于准备就去Google面试，结果被爆，与其说是能力问题，不如说是态度问题。

题目 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].思路分析：Pe

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.Return all such possible sentences.For example, given s = “catsanddo

<br />///这题不是很理解有待于重做，BFS+模拟，模拟有点麻烦 #include<stdio.h> #include<string.h> #include<queue> using namespace std; queue<int> q; int w,h; char map[45][45]; int dl[4][2]={{0,-1},{-1,0},{0,1},{1,0}}; int vis[45][45]; int dire,k;///指向

这是一道比较简单的DP，通过分析可以设最后拿走的牌为i，则所求的最优解就是i左边和右边子列的最小连乘积再加上x[a]*x[i]*x[b],应为i将原来的序列划分为两个子列，这两个子列符合“最有子结构”和“重叠子问题”的dp特点，他们的最优解互相之间没有影响，只会影响全局问题的最优解，在POJ discuss中的解析比较经典，摘录如下，以后做题可以常常看看discuss，就当学习，但是还是要独立思考为主对于整个牌的序列，最左端和最右端的牌是不能被取走的，除这两张以外的所有牌> ，必然有一张最后取走。取走这

路线可能有多条，线路要求输出的是按字典序搜索出现的第一个路线；也就是要从(1,1）开始；思路：从（1，1）点开始，直接深搜八个方向，注意方向的优先顺序；直接搜索就可以；编程要细心，如将==写成=导致调试一个多小时，此类错误以后务必杜绝！#include #include #include using namespace std;int map[30][30],c,r,num;//最大的行数、列数和已遍历点数string ans;//教训由于一个==写成=号而调试近一个小时！吸取教训，

理解算法的好方法可以单步调试查看关键变量的变化过程，如head和next，同时画出搜索树分析#include #include #define SIZE 100005using namespace std;//搜索是一种暴力的穷举//分1 - 1 * 2三个方向广搜queue x;bool vistied[SIZE];//记录点是否搜过int step[SIZE];//记录当前搜索的步数int bfs(int n,int k){ int head,next; x.

#include #include #include #define N 30#define f(i,a,b) for(i=a;i

POJ 1040TransportationTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 2872 Accepted: 1176DescriptionRuratania is just entering c

很经典的BFS搜索 走迷宫选取经过门最少的路线，这题POJ测试数据设计不全面，changeDir数组赋值错误也可以过。。。主要图的数据结构存储方式和算法实现参考了http://blog.youkuaiyun.com/bobten2008/article/details/5093307(1)首先是建图, 由于输入给的都是线段, 但是我们平常处理这类问题都是转换为网格来做的, 因此需要将线段转换为网格.转

这题和编程之美上面的“地板覆盖”问题有点像，不同的是，编程之美上面只需要判定能否覆盖，这题需要求出总方案数目结题报告转自 http://duanple.blog.163.com/blog/static/709717672008930104124684/题意：给你一个h*w的矩形，用一个1*2的小矩形去填充，问有多少种填充方法，不考虑对称性。 关键点提示： 1.DFS部分

这题主要求二叉树结点到根结点的路径长度，基本的思路是 比较a与b，如果a大则当前结点是左孩子，a-b作为父结点的左数，父结点的右数与当前右数相等；如果b大则当前结点为右孩子，同理可以求父结点，直到父结点为（1，1）遍历结束。当用原始的递归算法会超时，需要考虑a=1或b=1的特殊情况，同时利用a与b的倍数关系加快遍历速度Source CodeProblem: 2499

题目链接  http://cs.nyu.edu/courses/spring12/CSCI-GA.2560-001/prog1.html题目大意：给定n个任务的时间、价值及先后序关系，求一个可行的任务子集，使得时间之和不大于deadline，价值之和不小于targetVaule，且不可出现逆序。算法思路：题目已经给出算法，转化为状态空间搜索问题（tree-structured state

IDA*算法是A*算法和迭代加深ID算法的结合，先搜录相关资料如下[1] 搜索算法：IDA*算法,这是对IDA*算法的一个解释和学习笔记http://blog.youkuaiyun.com/urecvbnkuhBH_54245df/article/details/5856756[2] A*以及迭代加深的A*算法（IDA*） 两种算法的对比 http://www.cnblogs.com/semo/archive

作为AI课的预习笔记，搜索了下启发式搜索 A*算法的一些经典资料，整理到下面，这个算法很有意思，有时间自己实现一下。[1] 一个老外写的A*算法实例解说，写的非常翔实易懂，这是一个中文翻译版 http://blog.youkuaiyun.com/crayondeng/article/details/12342989[2] 这个基于[1]的阅读，添加了自己的理解性的注释 http://www.cnblogs.c

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:"((()))", "(()())", "(())()", "()(())", "()()()"思路分析：

The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially pos

Given a binary search tree, write a function kthSmallest to find the kth smallest element in it.Note: You may assume k is always valid, 1 ≤ k ≤ BST's total elements.Follow up:What if the BST is modifi