Elinx-CSDN博客

原创 LintCode Linked List专辑

1. Reverse Linked List2. Remove Duplicates from Sorted List3. Remove Duplicates from Sorted List II4. Reverse Linked List II5. Majority Number III6. Partition List7.1. Reverse Li...

2018-05-05 22:31:16 321

原创 LintCode二分查找专辑

1. First Position of Targetclass Solution {public: /** * @param nums: The integer array. * @param target: Target to find. * @return: The first position of target. Position star...

2018-04-13 21:37:00 253

原创枚举/组合

subsets descriptionclass Solution {public: /** * @param nums: A set of numbers * @return: A list of lists */ vector<vector<int>> subsets(vector<int> &a...

2018-04-07 12:33:23 265

原创 [DP] LeetCode Regular Expression Matching

algorithm procedureif the current text character is matching with the pattern character, which means the last character matched, then the result depends on the previous status: res(r,c)=res(r−1,c−1

2018-01-04 23:00:44 165

原创 LeetCode 2Sum, 3Sum, 4Sum

two sum problemthree accepted solutions are provided here with decreased time complexity.class Solution {public: vector<int> twoSum(vector<int>& nums, int target) { vector<int> res;#if 0

2018-01-03 22:29:41 176

原创 LeetCode 8-9 Solutions

8. String to Integer(atoi)class Solution {public: bool isDigit(char c) { return c >= '0' and c <='9'; } int myAtoi(string str) { size_t idx = 0; size_t len = str.length();

2018-01-01 21:55:12 172

原创 [DP] Hackerank The Coin Change Problem

You can find the problem description here, Sample Input 110 42 5 3 6Sample Output 15This is a typical knapsack problem with repetitions which can be easily solved by making a transaction table from to

2017-12-31 11:58:34 346

原创 Dynamic Programming part 1 -- Knapsack Problem

This article is adapted from this pdf.During a robbery, a burglar finds much more loot than he had expected and has to decide what to take. His bag (or “knapsack”) will hold a total weight of at most

2017-12-30 12:09:20 452

原创 [DP] LeetCode 5 Longest Palindromic Substring

O(N2)O(N^2) codeclass Solution {public: string longestPalindrome(string str) { int size = str.length(); int ps = 0, maxLen = 1; vector<vector<bool>> isP(size, vector<bool>(

2017-12-30 10:36:17 136

原创 [DP] Leetcode 300 Longest Increasing Subsequence

O(N2)O(N^2) codeclass Solution {public: int lengthOfLIS(vector<int>& seq) { if (seq.empty()) return 0; // lens indicate the sequence END by element t vector<int> lens(seq.s

2017-12-30 10:24:38 225

原创 1.3.1 FatMouse trade

WA:#include#include#includeusing namespace std;class Node {public: double ratio, mount,value; Node(double r,double m,double v){ ratio=r; mount=m; value=v; } bool operator>(const

2013-04-02 22:38:47 630

原创 1.2.8 Lowest Bit

#include#include#includeusing namespace std;int main(int argc, char* argv[]){ int get, count; while(cin >> get) { if(get == 0) return 1; bitset bs(get); count = 0; int len = bs

2013-04-02 21:10:47 609

原创 1.2.7 GPA

#include#include#includeusing namespace std;#define max 1000int main(int argc, char* argv[]){ double result; int i; char line[max];l: while(cin.getline(line, max)) { int len = string(

2013-04-01 21:53:26 938

原创 Huffman树

要解决的问题是关于霍夫曼树的，题目在：http://acm.hdu.edu.cn/showproblem.php?pid=1053直接写代码吧，不得不说写得真的很复杂，调试了半天，出了好多错哈：#include#include#include#includeusing namespace std;const int R = 27; // A~Z, _cons

2013-01-18 15:40:27 648

原创 PQ使用

priority_queue默认用 priority。第一次代码比较复杂：pq.cpp#include#includeusing namespace std;class NODE { public: char ch; int freq; bool operator>(const NODE n1) const

2013-01-17 16:42:10 1763 1

原创 DFS走迷宫

BFS走出的是一条最短路径，DFS走出的不一定是最短路径。只要把前面的BFS代码稍作修改，改成递归的形式即可。BFS使用的队列的数据结构来存储中间结果，DFS使用的是栈，因此DFS使用递归的形式。#include#include#include#include#define MAXLEN 100int edge_to[MAXLEN];int maze[MAXLEN][MA

2013-01-04 14:05:35 949

原创 BFS走迷宫

简单的走迷宫的算法，BFS就是把当前的顶点的邻接顶点都入队列，然后队列首端出队列，重复，知道所有的通路，然后一个一个的往回找即可。#include#include#include#include#define MAXLEN 100int edge_to[MAXLEN];int maze[MAXLEN][MAXLEN];bool visited[MAXLEN][MAXLEN];

2013-01-04 10:52:49 841