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;
}
}
};

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