LeetCode-Divide Two Integers

Divide two integers without using multiplication, division and mod operator.

思路是使用人做除法时的运算式 每次取前除数长度或加1位还原成数字，做减法求出该位的商，需要注意很多特殊情况。。

class Solution {
public:
    //a/b
    void Sub(string& a,string& b,string&ret,long long ib,stringstream& ss){
        int i=0;
        string a1;
        bool all=false;
        bool end=false;
        if(a.length()<b.length()){
            ret[0]='0';
            ret.resize(1);
            return;
        }
        while(i<b.length()-1){
            ret[i]='0';
            i++;
        }
        while(a.length()>=b.length()){
            bool big=true;
            for(int i=0;i<b.length();i++){
                if(a[i]<b[i]){
                    big=false;
                    break;
                }
                if(a[i]>b[i]){
                    break;
                }
            }
            if(!big){
                if(a.length()==b.length())
                {
                    break;
                }
                a1=a.substr(0,b.length()+1);
            }
            else{
                a1=a.substr(0,b.length());
            }
            ss.clear();
            ss<<a1;
            long long ia;
            ss>>ia;
            ss.clear();
            int count=0;
            while(ia>=ib){
                ia-=ib;
                count++;
            }
            int len=a1.length();
            if(ia==0){
                a1="";
                if(all)
                for(int j=0;j<len-1;j++){
                    ret[i]='0';
                    i++;
                }
                all=true;
            }
            else{
                all=false;
                ss<<ia;
                ss>>a1;
                ss.clear();
            }
            ss.clear();
            string a2=a.substr(len,a.length()-len);
            a1=a1+a2;
            if(a2=="")end=true;
            if(a1!=""){
                ss<<a1;
                ss>>ia;
                ss.clear();
                ss<<ia;
                ss>>a;
            }
            else a=a1;
            ret[i]=count+'0';
            i++;
        }
        if(!end){
            ret[i]='0';
            i++;
        }
        ret.resize(i);
    }
    int divide(int dividend, int divisor) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(divisor==1)return dividend;
        if(divisor==-1)return -dividend;
        if(dividend==0)return 0;
        if(divisor==0){
            if(dividend>0)return 2147493647;
            else return 0x80000000;
        }
        long long dvd=dividend,dvr=divisor;
        bool n1=false,n2=false;
        if(dvd<0){
            dvd=-dvd;
            n1=true;
        }
        if(dvr<0){
            dvr=-dvr;
            n2=true;
        }
        stringstream ss;
        string dd,dr;
        ss<<dvd;
        ss>>dd;
        ss.clear();
        ss<<dvr;
        ss>>dr;
        ss.clear();
        string result;
        result.resize(dd.length());
        Sub(dd,dr,result,dvr,ss);
        ss.clear();
        ss<<result;
        long long ret;
        ss>>ret;
        if(n1==n2){
            return (int)ret;
        }
        else{
            int ret2=-ret;
            return ret2;
        }
    }
};