Problem:

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

If it is overflow, return MAX_INT.

Solution:

不能乘除就加减就行了，但是一个问题是加减有可能速度太慢，因此需要转换，由于任何一个数都能表示成二进制，所以有dividend=divisor*（a*2^1 + b*2^2 + ...... + m*2^k）

所以只要计算出所有divisor*2^k，然后减去即可。

题目大意：

给定两个整数，要求不用乘除法和取模运算，计算出a/b的值，当结果越界的时候输出INT最大值

Java源代码（260ms）：

public class Solution {
    public int divide(int dividend, int divisor) {
        long a,b,flag=0,sum=0;
        long[] map=new long[33],times=new long[33];
        if(dividend<0 && divisor<0)flag=1;
        else if(dividend>0 && divisor>0)flag=1;
        a=Math.abs((long)dividend);
        b=Math.abs((long)divisor);
        int i=0;
        map[0]=b;times[0]=1;
        while(map[i]<=a){
            i++;
            map[i]=map[i-1]+map[i-1];
            times[i]=times[i-1]+times[i-1];
        }
        for(int j=i-1;j>=0;j--){
            while(a >= map[j]){
                a-=map[j];
                sum+=times[j];
            }
        }
        sum=flag==1?sum:-sum;
        if(sum>Integer.MAX_VALUE || sum<Integer.MIN_VALUE)return Integer.MAX_VALUE;
        return (int)sum;
    }
}

C语言源代码（4ms）：

long long ABS(long long a){
    return a>0?a:-a;
}
int divide(int dividend, int divisor) {
    int i=0,j,flag=0;
    long long sum=0,a,b,map[33],times[33],STOP=1;
    STOP=((long long)2147483647)+1;
    if(divisor==0)return INT_MAX;
    if(dividend==0)return 0;
    if((dividend>0 && divisor>0) || (dividend<0 && divisor<0))flag=1;
    a=ABS((long long)dividend);
    b=ABS((long long)divisor);
    map[0]=b;times[0]=1;
    while(map[i] <= a && i<33){
        i++;
        map[i]=map[i-1]+map[i-1];
        times[i]=times[i-1]+times[i-1];
    }
    for(j=i-1;j>=0;j--){
        while(a >= map[j]){
            a-=map[j];
            sum+=times[j];
        }
    }
    sum=flag?sum:-sum;
    if(sum<INT_MIN || sum > INT_MAX)return INT_MAX;
    return (int)sum;
}

C++源代码（24ms）：

class Solution {
public:
    int divide(int dividend, int divisor) {
        int i=0,j,flag=0;
        long long sum=0,a,b,map[33],times[33],STOP=1;
        STOP=STOP<<31;
        if(divisor==0)return INT_MAX;
        if((dividend>0 && divisor>0) || (dividend<0 && divisor<0))flag=1;
        a=abs((long long)dividend);
        b=abs((long long)divisor);
        map[0]=b;times[0]=1;
        while(map[i] <= STOP){
            i++;
            map[i]=map[i-1]+map[i-1];
            times[i]=times[i-1]+times[i-1];
        }
        for(j=i-1;j>=0;j--){
            while(a >= map[j]){
                a-=map[j];
                sum+=times[j];
            }
        }
        sum=flag?sum:-sum;
        if(sum<INT_MIN || sum > INT_MAX)return INT_MAX;
        return (int)sum;
    }
};

Python源代码（72ms）：

class Solution:
    # @param {integer} dividend
    # @param {integer} divisor
    # @return {integer}
    def divide(self, dividend, divisor):
        flag=0
        if dividend>0 and divisor>0:flag=1
        elif dividend<0 and divisor<0:flag=1
        if dividend==0:return 0
        dividend=abs(dividend)
        divisor=abs(divisor)
        map=[0 for i in range(33)]
        times=[1 for i in range(33)]
        i=0
        map[0]=divisor;times[0]=1
        while map[i]<=dividend:
            i+=1
            map[i]=map[i-1]+map[i-1]
            times[i]=times[i-1]+times[i-1]
        j=i-1;sum=0
        while j>=0:
            while map[j]<=dividend:
                dividend-=map[j]
                sum+=times[j]
            j-=1
        sum = -sum if flag==0 else sum
        if sum>2147483647:return 2147483647
        return sum