Leetcode 43. Multiply Strings

43. Multiply Strings

Total Accepted: 75018 Total Submissions: 300195 Difficulty: Medium

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 internal library such as BigInteger.

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 (M) Add Two Numbers (E) Plus One (E) Add Binary

准备某公司面试的时候碰到的面经题，刚好做过就看了看发上来了。

之前AC的代码很简洁，新建个数组，size是俩 string size之和

12 - length 2

13 - length 2

new int[4]

外层循环 i = 1, 0

内层循环 j = 1, 0

p[1 + 1 + 1] += 6, p[1 + 0 + 1] += 2

p[0 + 1 + 1]+ =3, p[0+ 0 + 1]+ = 1 - > p[0, 1, 5, 6]

然后从后向前（低到高位）处理进位，没有超过10的，所以直接从头开始粘贴num输出，156。

public class Solution {    public String multiply(String num1, String num2) {        int n1 = num1.length(), n2 = num2.length();        int[] products = new int[n1 + n2];        for (int i = n1 - 1; i >= 0; i--) {            for (int j = n2 - 1; j >= 0; j--) {                int d1 = num1.charAt(i) - '0';                int d2 = num2.charAt(j) - '0';                products[i + j + 1] += d1 * d2;            }        }        int carry = 0;        for (int i = products.length - 1; i >= 0; i--) {            int tmp = (products[i] + carry) % 10;            carry = (products[i] + carry) / 10;            products[i] = tmp;        }        StringBuilder sb = new StringBuilder();        for (int num : products) sb.append(num);        while (sb.length() != 0 && sb.charAt(0) == '0') sb.deleteCharAt(0);        return sb.length() == 0 ? "0" : sb.toString();    }}