Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2.
Note:
- The length of both num1 and num2 is < 110.
- Both num1 and num2 contains only digits 0-9.
- Both num1 and num2 does not contain any leading zero.
- You must not use any built-in BigInteger library or convert the inputs to integer directly.
class Solution {
public String multiply(String num1, String num2) {
int len1 = num1.length(), len2 = num2.length();
int[] mul = new int[len1+len2];
for (int i = len1-1; i >= 0; i--) {
for (int j = len2-1; j >= 0; j--) {
int p1 = i + j; //高位数字
int p2 = i + j + 1; //低位数字
int temp = (num1.charAt(i) - '0') * (num2.charAt(j) - '0');
int num = temp + mul[p2];
mul[p1] += num / 10;
mul[p2] = num % 10;
}
}
StringBuilder sb = new StringBuilder();
for (int i : mul) {
if (!(i == 0 && sb.length() == 0)) sb.append(i);
}
return sb.length() == 0 ? "0" : sb.toString();
}
}
- 按照正常的乘法逻辑,从右往做遍历每位数,i+j+1是低位,i+j是高位;
- 在遍历时,两个数的对应位的数相乘;
- 所得和的十位存入高位,个位存入低位。