leetcode_43

Golang:

思路：模拟大数乘法，实现麻烦，实现难度并不大

代码如下：

func multiply(num1 string, num2 string) string {
    if num1[0]=='0'||num2[0]=='0'{
        return "0"
    }
    bytes1:=[]byte(num1)
    bytes2:=[]byte(num2)
    res:="0"
    s:=""
    for i:=len(num2)-1;i>=0;i--{
        var temp strings.Builder
        temp.WriteString(s)
        s+="0"
        n:=0
        t:=int(bytes2[i]-'0')
        if t!=0{
            for j:=len(num1)-1;j>=0;j--{
                mu:=t*int(bytes1[j]-'0')+n
                temp.WriteString(strconv.Itoa(mu%10))
                n=mu/10
                if j==0&&n!=0{
                    temp.WriteString(strconv.Itoa(n))
                }
            }
        }
        res=plusStrings(reverseStrings(temp.String()),res)
    }
    return res
}

func plusStrings(a,b string) string{
    i,j:=len(a)-1,len(b)-1
    var res strings.Builder
    n:=0
    for i>=0&&j>=0 {
        t1, _ := strconv.Atoi(a[i : i+1])
        t2, _ := strconv.Atoi(b[j : j+1])
        t3 := t1 + t2 + n
        res.WriteString(strconv.Itoa(t3 % 10))
        n = t3 / 10
        i--
        j--
    }
    if i!=j{
        for i>=0{
            t1, _ := strconv.Atoi(a[i : i+1])
            t3 := t1 + n
            res.WriteString(strconv.Itoa(t3 % 10))
            n = t3 / 10
            i--
        }
        for j>=0{
            t1, _ := strconv.Atoi(b[j : j+1])
            t3 := t1 + n
            res.WriteString(strconv.Itoa(t3 % 10))
            n = t3 / 10
            j--
        }
    }
    if n!=0{
        res.WriteString(strconv.Itoa(n))
    }
    return reverseStrings(res.String())
}


func reverseStrings(s string) string{
    bytes:=[]byte(s)
    i,j:=0,len(s)-1
    for i<j{
        bytes[i],bytes[j]=bytes[j],bytes[i]
        i++
        j--
    }
    return string(bytes)
}

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