43. 字符串相乘 @Python @Go

原题

https://leetcode.cn/problems/multiply-strings/description/

思路

数组储存数字

复杂度

时间：O(m+n)
空间：O(m+n)

Python代码

class Solution:
    def multiply(self, num1: str, num2: str) -> str:
        if num1 == '0' or num2 == '0':
            return '0'
        ans = ''
        m, n = len(num1), len(num2)
        # 数组储存乘积的数字
        l = [0] * (m + n)
        for i in range(m -1, -1, -1):
            for j in range(n-1, -1, -1):
                multi = int(num1[i]) * int(num2[j])
                # 乘积的个位索引
                p1 = i+j+1
                # 乘积的进位索引
                p2 = i + j 
                # 加上原来位数的和
                sum = multi + l[p1]
                l[p1] = sum % 10
                l[p2] += sum // 10
        for digit in l:
            # 去掉前导0
            if len(ans) == 0 and digit == 0:
                continue 
            ans += str(digit)
        return ans 
       

Go代码

func multiply(num1 string, num2 string) string {
	if num1 == "0" || num2 == "0" {
		return "0"
	}
	var ans string
	m, n := len(num1), len(num2)
	// 切片储存乘积的数字
	l := make([]int, m+n)
	for i := m - 1; i > -1; i-- {
		for j := n - 1; j > -1; j-- {
			multi := int(num1[i]-'0') * int(num2[j]-'0')
			// 乘积的个位索引
			p1 := i + j + 1
			// 乘积的进位索引
			p2 := i + j
			// 加上原来位数的和
			sum := multi + l[p1]
			l[p1] = sum % 10
			l[p2] += sum / 10
		}
	}
	// 将数组转为字符串
	for _, digit := range l {
		// 去掉前导0
		if len(ans) == 0 && digit == 0 {
			continue
		}
		ans += string('0' + digit)
	}
	return ans
}