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RSS订阅原创 算法分析与设计第十九周： 567. Permutation in String
class Solution(object): def checkInclusion(self, s1, s2): l1 = [0 for _ in range(26)] l2 = [0 for _ in range(26)] size1 = len(s1) size2 = len(s2) if size2 <
2017-07-05 13:28:11
290
转载 Lua学习笔记之function type
Here we list all functions and types from the C API in alphabetical order. Each function has an indicator like this:[-o, +p, x]The first field, o, is how many elements the function pops from
2017-07-05 10:33:52
487
转载 Lua学习笔记之Coroutine
2.11 – CoroutinesLua supports coroutines, also called collaborative multithreading. A coroutine in Lua represents an independent thread of execution. Unlike threads in multithread systems, however
2017-07-04 20:59:26
398
转载 Lua学习笔记之Expressions
Both function calls and vararg expressions can result in multiple values. If an expression is used as a statement (only possible for function calls (see §2.4.6)), then its return list is adjusted to
2017-07-04 19:33:18
283
原创 课程调度问题：LeetCode 630. Course Schedule III
题目描述： There are n different online courses numbered from 1 to n. Each course has some duration(course length) t and closed on dth day. A course should be taken continuously for t days and must be fini
2017-06-26 14:19:47
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原创 算法分析与设计第十八周：592. Fraction Addition and Subtraction
class Solution(object): def fractionAddition(self, expression): def gcd(a, b): a = abs(a) b = abs(b) if a == 0 or b == 0: return 1
2017-06-26 10:24:45
363
原创 8.8 精确4SAT问题证明
思路： 若一个问题为NP问题，要证明该问题是NPC问题，只需要将一个已知的NPC问题规约为该问题即可。 规约过程分两步：将已知的NPC问题的实例转化为该问题的实例。将已知的NPC问题的解转化为该问题的解。已知3SAT问题是NPC问题，所以 利用3SAT问题进行规约证明。 证明过程：4精确SAT问题跟3SAT问题都是NP问题规约过程： 给定一个3SAT问题的实例：X={x1,x2,
2017-06-13 17:18:27
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原创 算法分析与设计第十七周：611. Valid Triangle Number
class Solution {public: int triangleNumber(vector<int>& nums) { //排序数组 sort(nums.begin(), nums.end()); //删除零元素 while (1) { //若数组中边长个数少于3，则直接返回0
2017-06-12 15:51:21
313
原创 算法分析与设计第十六周：582. kill process
class Solution(object): def killProcess(self, pid, ppid, kill): """ :type pid: List[int] :type ppid: List[int] :type kill: int :rtype: List[int] """
2017-06-05 15:40:24
300
原创 算法分析与设计第十五周：593. Valid Square
import mathclass Solution(object): def validSquare(self, p1, p2, p3, p4): #计算两点之间距离的函数 def dist(p1, p2): return math.sqrt((p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2)
2017-05-27 15:07:43
238
原创 算法分析与设计第十四周：424. Longest Repeating Character Replacement
class Solution(object): def characterReplacement(self, s, k): if s == "": return 0 char = [0 for _ in range(26)] i, j = 0, 0 maxCnt = 1 for c in s
2017-05-22 17:21:30
248
原创 算法分析与设计第十三周：582. Kill Process
class Solution(object): def killProcess(self, pid, ppid, kill): dict1 = {} size = len(pid) res = [kill] for i in range(size): if dict1.get(ppid[i]) == Non
2017-05-15 17:55:55
312
原创 算法分析与设计第十二周：473. Matchsticks to Square
Description: Remember the story of Little Match Girl? By now, you know exactly what matchsticks the little match girl has, please find out a way you can make one square by using up all those matchstic
2017-05-08 21:49:20
333
原创 【0-1 knapsack】 474. Ones and Zeroes
Description: In the computer world, use restricted resource you have to generate maximum benefit is what we always want to pursue.For now, suppose you are a dominator of m 0s and n 1s respectively. On
2017-05-03 16:50:51
212
原创 560. Subarray Sum Equals K
Description: Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.Solution:暴力列举所有可能的子连续序列和计算前n个序列的和，并储存与map中，再根据nums[0:i] - nu
2017-05-02 16:53:34
603
原创 算法分析与设计第十一周：416. Partition Equal Subset Sum
思路： 1.对数组进行排序并取得数组的总和 2.从大到小对数组进行筛选 3.只要有一种可行方案即返回Trueclass Solution(object): def canPartition(self, nums): if len(nums) <= 1: return False sum1 = sum(nums) nu
2017-05-01 23:30:07
210
原创 491. Increasing Subsequences
class Solution(object): def findSubsequences(self, nums): #先将list转成tuple #然后将tuple保存在set中 #并利用set来判断list是否已存在 self.__res = set() def rec(nums, l, cnt, val):
2017-04-25 21:13:31
166
原创 算法分析与设计第十周：384. Shuffle an Array
class Solution(object): def __init__(self, nums): """ :type nums: List[int] """ self.__l = nums self.__size = len(nums) def reset(self): """
2017-04-24 14:20:32
313
原创 算法分析与设计第九周：45. Jump Game II
#DP，从前往后扫描，依次更新到达每个位置所需要的步数class Solution(object): def jump(self, nums): size = len(nums) if size <= 1: return 0 steps = [size + 1 for _ in range(size)]
2017-04-17 22:14:33
189
原创 算法分析与设计第八周：300. Longest Increasing Subsequence
class Solution(object): def lengthOfLIS(self, nums): if nums == []: return 0 d = {1 : nums[0]} maxLength = 1 for num in nums[1:]: added = Fal
2017-04-10 17:40:47
181
原创 算法分析与设计第七周：134. Gas Station
题目简介： There are N gas stations along a circular route, where the amount of gas at station i is gas[i].You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to
2017-04-01 16:19:55
287
原创 算法分析与设计第六周：236. Lowest Common Ancestor of a Binary Tree
class Solution {public: TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { p_father.clear(); q_father.clear(); //为两个目标节点寻找father findFather(
2017-03-28 00:52:19
183
原创 算法分析与设计第五周习题：95. Unique Binary Search Trees II
与96题的对比： 95题是96题Unique Binary Search Trees的升级版，95题增加了构建出所有树的要求，而不仅仅是计算树的数量。对于96题，我们可以轻易得出树数量的规律。即：1.长度为3的树可由根和一个长度为2的子树构成（即以3为根或者以1为根） 或者 由根和两个长度为1的子树构成（ 即以2为根） 2. 长度为4的子树可由根和一个长度为3的子树构成（即以4
2017-03-21 20:04:39
348
原创 算法分析与设计第四周：513. Find Bottom Left Tree Value
class Solution(object): __maxLevel = 0 __maxLeft = 0 def findBottomLeftValue(self, root): level = 0 def rec(root, level): if root != None: level
2017-03-18 18:10:35
242
原创 算法分析与设计第三周习题：图论算法之310. Minimum Height Trees
今天刚上完算法课，课上知识内容包括拓扑排序和图论一些基本知识，所以我就去leetcode上面找了一道有关图论的题目来巩固自己所学的知识。题目简介： For a undirected graph with tree characteristics, we can choose any node as the root. The result graph is then a rooted tree.
2017-03-09 22:27:08
272
原创 算法分析与设计第二周习题：分治算法之P215,P169,P53
题目链接 题目名称：215. Kth Largest Element in an Array 题目描述： Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.Fo
2017-03-05 20:36:40
496
原创 分治算法之归并排序
归并排序分两步： 第一步：将数组递归分为两部分。 第二步：将两部分数组递归合并为一个数组。def mergeSort(l1): print('mergeSort:',l1) length = len(l1) if length == 1: return l1 mid = int(length / 2) l = [] merge(me
2017-03-01 21:31:31
243
原创 算法分析与设计第一周习题：2. Add Two Numbers
题目链接 题目名称：2. Add Two Numbers 题目描述： You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single di
2017-02-24 17:56:05
336
原创 memo
除以上较为普遍的算法以外，还有多种由协同过滤算法衍生出来的算法。利用均值平方差（Cacheda,Carneiro,Fernández,&For-moso,2011）进行协同过滤的算法，还有使用模糊集合论（fuzzy set theory）为不同的相似度赋权值的算法（Lu,Shambour,Xu,Lin,andZhang，2013）。还有学者将熵（entropy）、稀疏度（sparsity）、神经网
2016-04-13 15:28:36
256
原创 最小生成树--PRIM
#define MAXVALUE 100000//以1节点作为缺省开始搜索的节点void prim(int** graph, int vertices, int startPoint = 1) { int minEdge = MAXVALUE, foundV = 1; int length = vertices; bool* visited = new bool[lengt
2015-09-02 16:00:49
516
原创 最小生成树-Kruskal
#include <iostream>using namespace std;#define MAXVALUE 100000//顶点以1为最小点//保存数组中以0为最小点void kruskal(int** graph, int vertices) { int length = vertices, foundV = 0, run = 0; //set用于记录每个顶点属于哪个集合
2015-09-02 11:26:30
405
原创 最小二叉堆
最小二叉堆#include <iostream>#include <vector>using namespace std;//模板函数的实现与声明要写在一个文件中template <class comparable>class binaryMinHeap {public: binaryMinHeap(); void insert(const comparable &value)
2015-09-01 13:31:14
584
原创 AVL平衡二叉树
#include <iostream>#include <cmath>#include <algorithm>#include <time.h>using namespace std;#define DEFAULTVALUE (-1)struct avlNode { int height; int value; avlNode* left; avlNode* r
2015-08-28 18:02:36
290
原创 二叉搜索树
二叉搜索树#include <iostream>using namespace std;#define length 10#define errorValue -1int num;struct node { node* left; node* right; int value; node(int a = errorValue, node* n1 = NULL, no
2015-08-28 15:26:29
299
原创 关于线程同步的实现机制---busy waiting
第一种线程同步机制：只利用一个全局变量来判断是否有线程正在使用critical section，这样就有可能出现，两个线程同时判断当前没有线程正在使用critical section的情况，从而同时进入了critical section。 实际结果输出：偶尔出现加减交替的情况，即偶尔出现了两个线程同时进入critical section的情况。#include <stdio.h>#include
2015-08-20 16:07:00
8261
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