Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string. Example 1: Input: num1 = "2", num2 = "3" Output: "6" Example 2: Input: num1 = "123&…

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. 给出两个字符串,返回对应数字想乘后的字符串,由于这个字符串可能很大,所以不能采用一般的乘法,这里用的方法是模拟手工的乘法运算,算法 本身很简单,就是当时写的时候有些很小的细节搞错了,找了…

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-negative. 就是实现大数乘法.嘿嘿,我先用long long直接乘试试手感,试过了不行.所以还是要用字符串表示才行. 我是这样做的,先实现两个子函数,一个是实现一个数和一个字符串的数相乘的结果…

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. 要求:字符串表示的数字可能无穷大,并且非负. class Solution { private: vector<string> tempStrs; public: string add…

转载:43. Multiply Strings 题目描述 就是两个数相乘,输出结果,只不过数字很大很大,都是用 String 存储的.也就是传说中的大数相乘. 解法一 我们就模仿我们在纸上做乘法的过程写出一个算法. 个位乘个位,得出一个数,然后个位乘十位,全部乘完以后,就再用十位乘以各个位.然后百位乘以各个位,最后将每次得出的数相加.十位的结果要补 1 个 0 ,百位的结果要补两个 0 .相加的话我们可以直接用之前的大数相加.直接看代码吧. public String multiply(Stri…

http://www.cnblogs.com/TenosDoIt/p/3735309.html https://blog.csdn.net/fly_yr/article/details/48055617 class Solution { public: string multiply(string num1, string num2) { int len1 = num1.length(); int len2 = num2.length(); vector<); ; 第一轮乘出来的结果只有len1…

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. 这道题让我们求两个字符串数字的相乘,输入的两个数和返回的数都是以字符串格式储存的,这样做的原因可能是这样可以计算超大数相乘,可以不受int或long的数值范围的约束,那么我们该如何来计算…

Given two non-negative integers num1 and num2represented as strings, return the product of num1 and num2, also represented as a string. Example 1: Input: num1 = "2", num2 = "3" Output: "6" Example 2: Input: num1 = "123&q…

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. 题意:计算字符串对应数字的乘积.参考JustDoIT的博客. 思路:先用一维向量存放计算的中间值,中间值存放对应位置乘积的和(暂不考虑进位的情况):然后针对进位的情况,进行计算:注意要跳…

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string. Example 1: Input: num1 = "2", num2 = "3" Output: "6" Example 2: Input: num1 = "123&…