[LeetCode] 43. Multiply Strings

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string.

Example 1:

Input: num1 = "2", num2 = "3"
Output: "6"

Example 2:

Input: num1 = "123", num2 = "456"
Output: "56088"

Note:

1. The length of both num1 and num2 is < 110.
2. Both num1 and num2 contain only digits 0-9.
3. Both num1 and num2 do not contain any leading zero, except the number 0 itself.
4. You must not use any built-in BigInteger library or convert the inputs to integer directly.

题意很简单，给两个string做乘法。无需考虑一些比如string中夹杂字母的case，但是不允许将input convert成数字再做乘法。影子题415。需要注意的几个地方是

• 如果num1有i个数字，num2有j个数字，最后的结果res最多只有i + j个数字
• 细心观察之后就发现，num1[i] 和 num2[j] 的乘积只可能是一个两位数，对应的就是 res[i+j] 和 res[i+j+1] 这两个位置

https://leetcode-cn.com/problems/multiply-strings/solution/gao-pin-mian-shi-xi-lie-zi-fu-chuan-cheng-fa-by-la/

JavaScript实现

/**
 * @param {string} num1
 * @param {string} num2
 * @return {string}
 */
var multiply = function(num1, num2) {
    let m = num1.length;
    let n = num2.length;
    let pos = new Array(m + n).fill(0);

    for (let i = m - 1; i >= 0; i--) {
        for (let j = n - 1; j >= 0; j--) {
            let mul = (num1.charAt(i) - '0') * (num2.charAt(j) - '0');
            let p1 = i + j;
            let p2 = i + j + 1;
            let sum = mul + pos[p2];
            pos[p1] += Math.floor(sum / 10);
            pos[p2] = sum % 10;
        }
    }
    let res = '';
    for (let p of pos) {
        if (!(res.length === 0 && p === 0)) {
            res += p;
        }
    }
    return res.length === 0 ? '0' : res;
};

Java实现

class Solution {
    public String multiply(String num1, String num2) {
        int m = num1.length();
        int n = num2.length();
        int[] res = new int[m + n];
        for (int i = m - 1; i >= 0; i--) {
            for (int j = n - 1; j >= 0; j--) {
                int product = (num1.charAt(i) - '0') * (num2.charAt(j) - '0');
                int p1 = i + j;
                int p2 = i + j + 1;
                int sum = product + res[p2];
                res[p1] += sum / 10;
                res[p2] = sum % 10;
            }
        }
        StringBuilder sb = new StringBuilder();
        for (int p : res) {
            if (!(sb.length() == 0 && p == 0)) {
                sb.append(p);
            }
        }
        return sb.length() == 0 ? "0" : sb.toString();
    }
}

LeetCode 题目总结