Multiply Strings ---leetcode
发布时间:2021-03-01 22:45:19 所属栏目:大数据 来源:网络整理
导读: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. Converting the input string to integer is?NOT?allowed. You should?NOT?use interna
Given two numbers represented as strings,return multiplication of the numbers as a string. Note:
Subscribe?to see which companies asked this question 这是来自于leetcode上的一道算法题目。 算法思路:用char存储大数的每一位,然后控制每一位之间的相乘和进位即可。 我的代码如下: class Solution { public: string multiply(string num1,string num2) { const char *ch1 = num1.c_str(); const char *ch2 = num2.c_str(); const unsigned int l1=num1.length(); const unsigned int l2=num2.length(); const unsigned int lc = l1 +l2 +1; //cout<<str1.length()+str2.length()+1<<endl; char *ch3 = (char *)malloc((lc+1)*sizeof(char)); for (int i = 0; i < lc; i++) { ch3[i]='0'; } for (int i = l1 - 1; i >= 0 ; i--) { unsigned int start = (l1-i); int re = 0; for (int j = l2 - 1; j >= 0 ; j--) { int temp = (ch1[i]-48)*(ch2[j]-48) + re; int result = ch3[lc - (l1-i)-(l2-j)]+(temp%10) -48; re =temp/10; if (result>9) { re+=result/10; result=result%10; } ch3[lc - (l1-i)-(l2-j)]= result+48; if (j==0) { ch3[lc - (l1-i)-(l2-j)-1]=re+48; } } } string a(ch3); free(ch3); int i=0; for(;i<lc-2;i++){ if(a.substr(i,1)=="0")continue; else break; } return a.substr(i,lc-i-1); } };结果: (编辑:上海站长网) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |