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[LeetCode] Multiply Strings
Multiply Strings Given two numbers represented as strings, return multiplication of the numbers as a string. Note: The numbers can be arbitrarily large and are non-negative. 解题思路: 按照小学的做法,每位...
分类:其他好文   时间:2015-05-06 21:16:09    阅读次数:118
DLL文件编译出错_DllMain@12 already defined in XXX.obj
编译MFC DLL程序时出现 Error 2 error LNK1169: one or more multiply defined symbols found E:\C++\HookTest\CopyFileDetour\CFDetour\Debug\1t.dll 1 1t 1 Error 1 error LNK2005: _DllMain@12 already defined...
分类:其他好文   时间:2015-05-04 20:19:05    阅读次数:265
[LeetCode]Multiply Strings
Given two numbers represented as strings, return multiplication of the numbers as a string. Note: The numbers can be arbitrarily large and are non-negative. 给定两个字符串表示的数,要求求这两个数的乘积,乘积用字符串表示。其中,字符串表...
分类:其他好文   时间:2015-04-24 22:47:50    阅读次数:148
LeetCode: Multiply Strings
Title:Given two numbers represented as strings, return multiplication of the numbers as a string.Note: The numbers can be arbitrarily large and are no...
分类:其他好文   时间:2015-04-24 16:10:24    阅读次数:97
js中闭包来实现bind函数的一段代码的分析
今天研究了一下bind函数,发现apply和call还可以有这样的妙用,顺便巩固复习了闭包。 1 var first_object = { 2 num: 42 3 }; 4 var second_object = { 5 num: 24 6 }; 7 function multiply(...
分类:Web程序   时间:2015-04-24 15:33:58    阅读次数:166
LeetCode (19) Multiply Strings
题目描述Given two numbers represented as strings, return multiplication of the numbers as a string.Note: The numbers can be arbitrarily large and are non-negative.对两组非负数字进行相乘,使用数组表示数字,且题目中说明数组很大,因此,因此不能直接将...
分类:其他好文   时间:2015-04-22 18:29:04    阅读次数:103
HDU 3074 Multiply game
Multiply gameTime Limit: 1000msMemory Limit: 32768KBThis problem will be judged onHDU. Original ID:307464-bit integer IO format:%I64d Java class name:...
分类:其他好文   时间:2015-04-18 17:33:58    阅读次数:131
43. Multiply Strings
题目:Given two numbers represented as strings, return multiplication of the numbers as a string.Note: The numbers can be arbitrarily large and are non-n...
分类:其他好文   时间:2015-04-17 23:42:10    阅读次数:130
Multiply Strings
public class Solution { public String multiply(String num1, String num2) { if(num1==null || num2==null || num1.length()*num2.length()==0) re...
分类:其他好文   时间:2015-04-16 06:43:13    阅读次数:94
opengl:凸包算法
准备工作判断点在有向线段的左侧可以通过叉积判断,如下为k在有向线段ab的左侧代码描述:double multiply(Point a, Point b, Point k) { double x1 = b.x-a.x; double y1 = b.y-a.y; double x2 = k.x-a.x; double y2 = k.y-a.y; return x1...
分类:编程语言   时间:2015-04-15 09:42:42    阅读次数:291
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