题目描述: 解题思路: java代码: public class LeetCode43 { public static void main(String[] args) { String num1="123"; String num2="45"; System.out.println(num1+"和"+num2+"相乘的结果是:"+new Solution().multiply(num1, num2)); } } class…

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.     各个位相乘,保存在数组中,最后再处理进位. 如 123*456     4,5,6      8,10,12         12,15,18…

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. 思路:直观思路,就是模拟乘法过程.注意进位.我写的比较繁琐. string multiply(string num1, string num2) { vector<int> v1, v…

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: 模拟乘法运算,可以完全按照模拟的思路,用num1的每一位乘num2,得到的数组按位保存在结果字符串中,并不断更新. 先把字符串反转,在逻辑上思路会更加清晰,当然不反转也可以. c…

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. 题目意思: 给两个string,计算string的乘积. string中的数可以非常大并且是非负数. (就是大数的乘法运算嘛...) 解题思路: 由于之…

Power Strings Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non…

Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative inte…

题目描述: 解决思路: 此题较简单,和前面[LeetCode67]方法一样. Java代码: public class LeetCode415 { public static void main(String[] args) { String a="1",b="9"; System.out.println(a+"和"+b+"相加的结果是:"+new Solution().addStrings(a, b)); } } class…

[题意]: 给出s串出来,能否找到一个前缀 ,通过多次前缀进行拼接.构成s串.如果有多个,请输出最多次数那个. 如:aaaa 可以用1个a,进行4次拼接 可以用2个a,进行2次拼接 可以用4个a,进行1次拼接 提供两种做法: 第一种是:利用kmp算法中求解  longest prefix table的方法, 找到最后一个位置的上的 p[len-1],如果用总长度减去(len-p[len-1])能否通过这一个前缀来拼接出来. 就是判断是否能整除. #include<bits/stdc++.h>…

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&…