leetcode:29. Divide Two Integers

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

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 r = mdiv(labs(dividend), labs(divisor));
        return r.first * minus;
    }

    pair mdiv(long de, long di) {
        if(de < di) return make_pair(0, de);
        pair 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 integer
def 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)