给定两个以字符串形式表示的非负整数 num1 和 num2，返回 num1 和 num2 的乘积，它们的乘积也表示为字符串形式。
注意：不能使用任何内置的 BigInteger 库或直接将输入转换为整数。
示例 1:
输入: num1 = "2", num2 = "3"
输出: "6"示例 2:
输入: num1 = "123", num2 = "456"
输出: "56088"提示：
- 1 <= num1.length, num2.length <= 200
- num1 和 num2 只能由数字组成。
- num1 和 num2 都不包含任何前导零，除了数字0本身。
解题：
按照竖式乘法分开来计算后加起来
func multiply(num1 string, num2 string) string {
if num1 == "0" || num2 == "0" {
return "0"
}
l := len(num2)
strNum := make([]string, 0)
s := ""
for i := l - 1; i >= 0; i-- {
str := strMulti(num1, num2[i]) + s
strNum = append(strNum, str)
s += "0"
}
s = "0"
for _, v := range strNum {
s = strAdd(s, v)
}
return s
}
func strMulti(num1 string, num uint8) string {
if num == '0' {
return "0"
}
s := ""
l := len(num1)
var a uint8 = 0
num -= '0'
for i := l - 1; i >= 0; i-- {
s = strconv.Itoa(int(((num1[i]-'0')*num+a)%10)) + s
a = ((num1[i]-'0')*num + a) / 10
}
if a > 0 {
s = strconv.Itoa(int(a)) + s
}
return s
}
func strAdd(a, b string) string {
if b == "0" {
return a
}
l := len(b)
if l > len(a) {
l = len(a)
a, b = b, a
}
var c uint8 = 0
s := ""
for i, j := len(a)-1, l-1; i >= 0; i-- {
if j >= 0 {
s = strconv.Itoa(int(a[i]+b[j]-2*'0'+c)%10) + s
c = (a[i] + b[j] - 2*'0' + c) / 10
} else {
s = strconv.Itoa(int(a[i]-'0'+c)%10) + s
c = (a[i] - '0' + c) / 10
}
j--
}
if c > 0 {
s = strconv.Itoa(int(c)) + s
}
return s
}官方解答：
1.做加法
func multiply(num1 string, num2 string) string {
if num1 == "0" || num2 == "0" {
return "0"
}
ans := "0"
m, n := len(num1), len(num2)
for i := n - 1; i >= 0; i-- {
curr := ""
add := 0
for j := n - 1; j > i; j-- {
curr += "0"
}
y := int(num2[i] - '0')
for j := m - 1; j >= 0; j-- {
x := int(num1[j] - '0')
product := x*y + add
curr = strconv.Itoa(product%10) + curr
add = product / 10
}
for ; add != 0; add /= 10 {
curr = strconv.Itoa(add%10) + curr
}
ans = addStrings(ans, curr)
}
return ans
}
func addStrings(num1, num2 string) string {
i, j := len(num1)-1, len(num2)-1
add := 0
ans := ""
for ; i >= 0 || j >= 0 || add != 0; i, j = i-1, j-1 {
x, y := 0, 0
if i >= 0 {
x = int(num1[i] - '0')
}
if j >= 0 {
y = int(num2[j] - '0')
}
result := x + y + add
ans = strconv.Itoa(result%10) + ans
add = result / 10
}
return ans
}2.做乘法
func multiply(num1 string, num2 string) string {
if num1 == "0" || num2 == "0" {
return "0"
}
m, n := len(num1), len(num2)
ansArr := make([]int, m+n)
for i := m - 1; i >= 0; i-- {
x := int(num1[i]) - '0'
for j := n - 1; j >= 0; j-- {
y := int(num2[j] - '0')
ansArr[i+j+1] += x * y
}
}
for i := m + n - 1; i > 0; i-- {
ansArr[i-1] += ansArr[i] / 10
ansArr[i] %= 10
}
ans := ""
idx := 0
if ansArr[0] == 0 {
idx = 1
}
for ; idx < m+n; idx++ {
ans += strconv.Itoa(ansArr[idx])
}
return ans
}