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：
大整数乘法，转换成大整数加法来算。
代码：
1 string add(string num1, string num2){ 2 int l1 = num1.length(), l2 = num2.length(); 3 if(l1 > l2) 4 return add(num2, num1); 5 int i; 6 int carry = 0; 7 string result = ""; 8 for(i = l1-1; i >= 0; i--){ 9 int tmp = num1[i] - '0' + num2[i+l2-l1] - '0' + carry; 10 result = (char)(tmp%10 + '0') + result; 11 carry = tmp/10; 12 } 13 for(i = l2-l1-1; i >= 0; i--){ 14 int tmp = num2[i] - '0' + carry; 15 result = (char)(tmp%10 + '0') + result; 16 carry = tmp/10; 17 } 18 if(carry > 0) 19 result = (char)(carry + '0') + result; 20 return result; 21 } 22 string multiply(string num1, string num2) { 23 // IMPORTANT: Please reset any member data you declared, as 24 // the same Solution instance will be reused for each test case. 25 int l1 = num1.length(), l2 = num2.length(); 26 if(l1 < l2) 27 return multiply(num2, num1); 28 if(l1 == 0) 29 return ""; 30 if(l1 == 1){ 31 if(num1[0] == '0') 32 return "0"; 33 if(num1[0] == '1') 34 return num2; 35 } 36 int i,j; 37 string result = ""; 38 for(i = l2-1; i >= 0; i--){ 39 string tstring = ""; 40 int carry = 0; 41 if(num2[i] == '0') 42 continue; 43 for(j = l1-1; j >= 0; j--){ 44 int tmp = (num2[i]-'0')*(num1[j]-'0') + carry; 45 tstring = (char)(tmp%10 + '0') + tstring; 46 carry = tmp/10; 47 } 48 if(carry > 0) 49 tstring = (char)(carry + '0') + tstring; 50 for(int k = 0; k < l2-i-1; k++) 51 tstring += '0'; 52 if(result == "") 53 result = tstring; 54 else 55 result = add(result, tstring); 56 } 57 if(result == "") 58 return "0"; 59 return result; 60 }
第一遍的方法比较麻烦，其实还可以先用int数组存中间结果，然后再转换为字符串。代码如下：
1 string multiply(string num1, string num2) { 2 if(num1 == "0" || num2 == "0") 3 return "0"; 4 int l1 = num1.length(), l2 = num2.length(); 5 int *n1 = new int[l1], *n2 = new int[l2], *r = new int[l1+l2]; 6 int i,j; 7 string result = ""; 8 for(i = 0; i < l1; i++) 9 n1[i] = num1[i]-'0'; 10 for(i = 0; i < l2; i++) 11 n2[i] = num2[i]-'0'; 12 memset(r, 0, sizeof(int)*(l1+l2)); 13 for(i = 0; i < l1; i++){ 14 for(j = 0; j < l2; j++){ 15 r[i+j+1] += n1[i]*n2[j]; 16 } 17 } 18 for(i = l1+l2-1; i >= 1; i--){ 19 result = (char)(r[i]%10+'0')+result; 20 r[i-1] += (r[i]/10); 21 } 22 if(r[0] > 0) 23 result = (char)(r[0]+'0')+result; 24 return result; 25 }