zem_nezer的博客_算法

POJ3050(DFS)

一定要根据数据规模想问题，这题数据规模就不大，时间复杂度大概是O(R*C*dir^step) = O(5*5*4^6)。规模不大，可以无脑DFS,还

算法

ACM

#include

i++

#define

原创 2023-08-23 09:53:20 48 阅读

POJ1979(Red and Black, DFS)

注意char处理, when need to manipluate inputs, take care of previous chars, such as ‘\n’ #define LOCAL#include #include har map[SIZE][SIZE];int dir

ACM

算法

i++

#include

#define

原创 2023-08-23 09:53:17 62 阅读

leetcode-1-two sum(brute force, hash table)

We can use brute force to get it, use two for loop i and j, which i = 1:n and j = i:n. However, the time complexity is O(n^2), which is no

ci

原创 2023-08-23 09:53:09 63 阅读

AOJ0558(cheese, BFS)

typical BFS usage: first use 2 array, one for record map, the other for shortest path/flag visit.BFS feature: the shortest path from one to the oth

ACM

算法

i++

#include

#define

原创 2023-08-23 09:52:39 48 阅读

POJ3669（meteor shower, BFS）TLE

when initial, try to convenient for later operation, e.g: if step[][] = -1, will hard to compare mini with other. However, my code still TLE for un

ACM

算法

#include

#define

i++

转载 2023-08-23 09:52:21 40 阅读

AOJ0525(bitset, 穷举)

这题有3点要注意： 1. the flip order is not related to result. 2. why we can simply count to maximum of number each column? I

算法

ACM

#include

#define

i++

原创 2023-08-23 09:52:15 50 阅读

POJ3069(贪心)

贪心，自己太想当然了，直接 left + 2 * R, 这当然是不对的。。你咋知道left+R 一定有点呢。。//#define LOCAL#include <cstdio>#include <cstring>#include <algorithm>#define N 1000using namespace std;int main(voi..

ACM

算法

#include

#define

i++

原创 2023-08-23 09:52:09 52 阅读

POJ2376(贪心)

参考：还是之前的技巧没有弄熟，在遇到这种直到满足某一条件时再进行改变时，就应该要想到用while, 方便而且思路清晰。从这里可以看出逻辑还是不够清楚，边界条件还是拿捏的不好。//#define LOCAL #include <cstdio> #include <cstring> #include <algorithm> #define N 25010 usi

算法

ACM

#include

#define

#ifdef

原创 2023-08-23 09:52:05 58 阅读

POJ3617(贪心)

没啥好说的，书上原题，比较坑就是输出需要每80个换行。//#define LOCAL#include #include #include #define SIZE 2000using namespace std;int ma

算法

ACM

#include

#define

i++

原创 2023-08-23 09:51:56 53 阅读

POJ2393(贪心)

水题，compare current storing value and the price of that day, if cheaper then assign, if not then update the store value.//#dg>#include <...

算法

ACM

#include

i++

#endif

原创 2023-08-23 09:51:53 45 阅读

POJ3253(贪心)

1.要考虑到规模为20，000 累加起来肯定会超的，要用long long 2. 思想就是先从正着推，一定是先切掉最长的那块，这样之后都不会受影响;再反着来想，就是先合并最小的//#define LOCAL#include <cstdio>#include <cstring>#include <algorithm>#include <que...

算法

ACM

#include

#define

#ifdef

原创 2023-08-23 09:51:50 53 阅读

POJ31900(贪心)

想对了一半，还是不扎实原本想将初始化和之后处理一起放到for里面的(i.e. 将push, ans = 1等放到for里面)，发现比较麻

算法

ACM

i++

#include

#ifdef

原创 2023-08-23 09:51:47 51 阅读

leetcode-3-Longest Substring Without Repeating Characters

Use sliding windows to solve the problem.First, we use the hash map to store how many character and which character wt hash map is valid or not. If valid, let ...

系统

原创 2023-08-23 09:51:43 35 阅读

leetcode-2-Add Two Numbers

The algorithm is straight-forward: Because the list represent reversely, we just need to add two elements then assign the residual to next element.However, I

系统

原创 2023-08-23 09:50:54 34 阅读

leetcode-8-String to Integer (atoi)

Follow the problem then we can get the answer. But from here I can see my coding ability is not we

系统

原创 2023-08-23 09:50:47 51 阅读

leetcode-4-Median of Two Sorted Arrays

弃疗，不大懂

系统

原创 2023-08-23 09:50:43 50 阅读

leetcode-7-Reverse Integer

straight forward, just keep in mind that we need to deal with overflow problem.

系统

原创 2023-08-23 09:50:36 45 阅读

leetcode-6-ZigZag Conversion

Use a trick, actually we just follow the zigzag move our current position, but we can use string to make a trick inponding t...

系统

原创 2023-08-23 09:50:33 42 阅读

leetcode-5-Longest Palindromic Substring

I am so stupid!!! Spent long time on that and cannot fix the bugs in short time.It has two solutions: the first is treat ea

系统

原创 2023-08-23 09:50:29 57 阅读

leetcode-9-Palindrome Number

Easy problem, first is convert number to string, then compare the string pairwise.

系统

原创 2023-08-23 09:50:25 49 阅读

leetcode-11-Container With Most Water

At first I thought I use the greedy: the move left pointer if the new one has longer height which can counteract the width

系统

原创 2023-08-23 09:50:20 76 阅读

leetcode-49-Group Anagrams

The problem is very tricky, first I use two rules:1. If two words are anagrams then they have same length2. If two words are anagrams then if have same mapThen, use loop to traverse the map and do ...

sed

原创 2023-08-23 09:50:08 58 阅读

leetcode-13-Roman to Integer

Though leetcode mark is as easy, but I think it is thicker than I though: we need to traverse it inversely, and when current roman is small than previous one, mins current roman, otherwise add it.

系统

原创 2023-08-23 09:49:11 41 阅读

leetcode-20-Valid Parentheses

Do it one hundred times, when meet left part of bracket, push it, right part check it. And finally check the stack is empty or not.

系统

原创 2023-08-23 09:49:07 40 阅读

leetcode-17-Letter Combinations of a Phone Number

It is like my previous problem, I do it many times, but still a stupid, cannot find a job. Use recursive, every store it

系统

原创 2023-08-23 09:29:23 71 阅读

leetcode-19-Remove Nth Node From End of List

Use a common trick call slow, fast pointer. Because the last n means the distance between last n and the end is n, so we create two pointer, thond is fast, and make fast to n s...

系统

原创 2023-08-23 09:29:17 66 阅读

leetcode-15-3Sum

Block me too much time, it is use two pointer to do. First is fix one numbers, then use two pointer to scan the remain sortedt remember to d...

系统

原创 2023-08-23 09:29:09 51 阅读

leetcode-21-Merge Two Sorted Lists

Compare the current value of each list, the assign the smaller node to merged list, then return it.It is straight forward, just need to know that i

系统

原创 2023-08-23 09:29:02 57 阅读

leetcode-22-Generate Parentheses

We can use recursive to solve, every time we can check if current situation can append ‘(’ or ‘)’. Then append it recursive. If meet left == n an

系统

原创 2023-08-23 09:28:53 49 阅读

leetcode-33-Search in Rotated Sorted Array

Use the most understandable method, in binary search, we can know that at least one part is sort in order. Base on that, we just neif current target is in that par...

系统

原创 2023-08-23 09:28:49 74 阅读

官方博客	全部文章	热门标签	班级博客
了解我们	网站地图	意见反馈

鸿蒙开发者社区	51CTO学堂
51CTO	软考资讯

51CTO博客