Elinx的专栏

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

1. Reverse Linked List 2. Remove Duplicates from Sorted List 3. Remove Duplicates from Sorted List II 4. Reverse Linked List II 5. Majority Number III 6. Partition List 7. 1. Reverse Li...

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

subsets description class Solution { public: /** * @param nums: A set of numbers * @return: A list of lists */ vector&lt;vector&lt;int&gt;&gt; subsets(vector&lt;int&gt; &a...

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

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>(

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

algorithm procedure if 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

You can find the problem description here, Sample Input 110 4 2 5 3 6Sample Output 15This is a typical knapsack problem with repetitions which can be easily solved by making a transaction table from to

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 100 int edge_to[MAXLEN]; int maze[MAXLEN][MAXLEN]; bool visited[MAXLEN][MAXLEN];

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 100 int edge_to[MAXLEN]; int maze[MAXLEN][MA

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

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 #include using namespace std; const int R = 27; // A~Z, _ cons

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

#include #include #include using 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

WA: #include #include #include using 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

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();