LeetCode 29 Divide Two Integers (C,C++,Java,Python)

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