43. 字符串相乘

目录

• 题目描述
• 思路
• 代码实现

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.

思路

模拟乘法的运算过程,两个非负数相乘，结果的位数至多为这两个非负数位数之和.不妨设这两个非负数位数分别为x,y，取这两个位数的最大的两个数相乘即(10^x-1)(10^y-1)=10^(x+y)-10^x-10^y+1,而10^(x+y)为最小的x+y+1位数,所以结果为一个x+y位数.

代码实现

class Solution {
public:
	string multiply(string num1, string num2) {
        
        if(num1.length()==0 || num2.length()==0)
        	return "";
        string ret(num1.size()+num2.size(),'0');
        for(int i=num2.size()-1;i>=0;--i) 
        {   
            int carry = 0;
            //位置对应关系利用平移不变性
            //它最初位于ret中的位置为：num1.size()+num2.size()-1-(num2.size()-1-i)
            int k  = num1.size()+num2.size()-1-(num2.size()-1-i);//初始化指针指向结果中的位置
            for(int j=num1.size()-1;j>=0;--j) 
            {
                int sum = (num2[i]-'0')*(num1[j]-'0')+(ret[k]-'0')+carry;
                ret[k] = sum%10 + '0';
                carry = sum/10;
                k--;
            }
            while(carry)
            {
                int sum = (ret[k]-'0')+carry;
                ret[k] = sum%10 + '0';
                carry = sum/10;
                k--;
            
            }   
        }
        int i = 0;
        while(i<ret.size() && ret[i]=='0') ++i;
        return i==ret.size()?"0":ret.substr(i);
    }          //例如123456*0=0
};