leetcode:29. Divide Two Integers

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描述

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

If it is overflow, return MAX_INT.

思路

这里我一开始的思路是通过减法来实现，不过实现了，超时了。实在想不到什么方法了，参考了一下讨论区，找到大神的思路。使用<<，>> 操作来运算。我在原来的代码基础上，将符号位的判断修改了一下，时间少了。

代码

class Solution {public:    int divide(int dividend, int divisor) {        if (!divisor || (dividend == INT_MIN && divisor == -1))            return INT_MAX;        int sign =(dividend > 0 && divisor > 0) || (dividend < 0 && divisor < 0) ? 1 : -1;        long long dvd = labs(dividend);        long long dvs = labs(divisor);        int res = 0;        while (dvd >= dvs) {             long long temp = dvs, multiple = 1;            while (dvd >= (temp << 1)) {                temp <<= 1;                multiple <<= 1;            }            dvd -= temp;            res += multiple;        }        return sign == 1 ? res : -res;     }};

结果

这里写图片描述

他山之玉

C++ O(n) solutioin

 int divide(int dividend, int divisor) {        if(!divisor || (dividend == INT_MIN && divisor == -1))            return INT_MAX;        int minus = ((dividend ^ divisor) >> 31) ? -1 : 1;        pair<long, long> r = mdiv(labs(dividend), labs(divisor));        return r.first * minus;    }    pair<long, long> mdiv(long de, long di) {        if(de < di) return make_pair(0, de);        pair<long, long> sub = mdiv(de >> 1, di);        long f, s;        f = (sub.first) << 1;        s = (sub.second << 1) + (de & 1);        if(s >= di) {            s -= di;            f += 1;        }        return make_pair(f, s);    }

Java O(n) solution

public class Solution {    public int divide(int dividend, int divisor) {        int sign=(dividend>0&&divisor>0)||(dividend<0&&divisor<0)?1:-1;        long divd=Math.abs((long)dividend),divi=Math.abs((long)divisor);        long res=0,lo=1,hi=divd;        while(lo<=hi){            long mid=lo+(hi-lo)/2;            if(mid*divi<=divd){                res=mid;                lo=mid+1;            }else{                hi=mid-1;            }        }        return res*sign==2147483648L?Integer.MAX_VALUE:(int)res*sign;    }}

python O(n) solution

class Solution:# @return an integerdef divide(self, dividend, divisor):    positive = (dividend < 0) is (divisor < 0)    dividend, divisor = abs(dividend), abs(divisor)    res = 0    while dividend >= divisor:        temp, i = divisor, 1        while dividend >= temp:            dividend -= temp            res += i            i <<= 1            temp <<= 1    if not positive:        res = -res    return min(max(-2147483648, res), 2147483647)

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