zoukankan      html  css  js  c++  java
  • [Leetcode 90] 43 Multiply Strings

    Problem:

    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.

    Analysis:

    It's the big number simulation problem. Recall the process how we compute the multiplication manually. Multiply the tow numbers digit by digit and sum the result up. Use a char array to store the result reversely and after the computing reverse it to the normal order. Pay attention of the carry bit and the space allocated to the result array (l1+l2+1 is definitely enough but l1+l2 is not). Also, if one of the given number is 0, just return 0 and don't bother to compute it even.

    Code:

     1 class Solution {
     2 public:
     3    string multiply(string num1, string num2) {
     4     // Start typing your C/C++ solution below
     5     // DO NOT write int main() function
     6     if (num1 == "0" || num2 == "0")
     7         return "0";
     8     
     9     int l1 = num1.length(), l2 = num2.length(), idx, cry;
    10     char *res = new char[l1+l2+1];
    11 
    12     for (int i=0; i<l1+l2; i++)
    13         res[i] = '0';
    14 
    15     for (int i=l1-1, inc=0; i>=0; i--, inc++) {
    16         idx = inc, cry = 0;
    17         for (int j=l2-1; j>=0; j--) {
    18             int mul = (num1[i]-'0') * (num2[j]-'0') + cry + (res[idx]-'0');
    19 
    20             res[idx] = ( mul % 10 ) + '0';
    21             cry = mul / 10;
    22             idx++;
    23         }
    24 
    25         if (cry > 0)
    26             res[idx] = cry + '0';
    27     }
    28 
    29     if (cry > 0)
    30         res[idx++] = cry + '0';
    31 
    32     res[idx] = '';
    33 
    34     for (int i=0, j=idx-1; i < j; i++, j--) {
    35         char t = res[i];
    36         res[i] = res[j];
    37         res[j] = t;
    38     }
    39 
    40     string r(res);
    41 
    42     return r;
    43 }
    44 };
    View Code
  • 相关阅读:
    反射
    IDEA配置数据库
    配置idea的maven镜像为aliyun
    蓝桥---芯片测试(思维)
    汉诺塔(思维、DP思想)
    立方数(质因子、优化)
    碎碎念(DP)
    牛牛战队的比赛地(三分)
    子段乘积(尺取、逆元)
    子段异或(位运算)
  • 原文地址:https://www.cnblogs.com/freeneng/p/3218928.html
Copyright © 2011-2022 走看看