LeetCode 29 Divide Two Integers（两个整数相除）（*）

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翻译

不用乘法、除法、取余操作，将两个数相除。如果它溢出了，返回MAX_INT

原文

Divide two integers without using multiplication, division and mod operator.If it is overflow, return MAX_INT.

代码

一心扑到了递归上，可惜没能写出来…………烦躁至极还是找了别人的答案……

class Solution {public:    int divide(int dividend, int divisor) {        if(!divisor) return INT_MAX;        if(divisor == 1) return dividend;        if(divisor == -1) {            if(dividend == INT_MIN) return INT_MAX;            else return -dividend;        }        bool s1 = dividend < 0;        bool s2 = divisor < 0;        unsigned int nom = s1 ? -dividend : dividend;        unsigned int den = s2 ? -divisor : divisor;        unsigned int rem = 0;        unsigned int quot = 0;        for(int i = 31; i >= 0; --i) {            rem <<= 1;            rem |= (nom >> i) & 1;            if(rem >= den) {                rem -= den;                quot |= (1 << i);            }        }        return s1^s2? -quot : quot;    }};

再来两个代码……（惭愧……）

public class Solution{    public int Divide(int dividend, int divisor)    {        //1. check overflow: 2 ways of over flow 1) 0 divisor; 2) int.Minvalue/(-1)        if (divisor == 0 || dividend == int.MinValue && divisor == -1) return int.MaxValue;        //2. calculate sign        int sign = dividend > 0 ^ divisor > 0 ? -1 : 1, result = 0;        long m = Math.Abs((long)dividend), n = Math.Abs((long)divisor);        //3. looping from 1 to possible maximum pow(2, x) to add into result        while (m >= n)        {            long subN = n;            for (int subCount = 1; m >= subN; subCount <<= 1, subN <<= 1)            {                m -= subN;                result += subCount;            }        }        return result * sign;    }}

public class Solution{    public int Divide(int dividend, int divisor)    {        if (divisor == 1) return dividend;        if (dividend == int.MinValue && divisor == -1 || divisor == 0) return int.MaxValue;        int sign = (dividend < 0) ^ (divisor < 0) ? -1 : 1, result = 0;        long dvd = Math.Abs((long)dividend), dvs = Math.Abs((long)divisor);        while (dvd >= dvs)        {            long sub = dvs;            int subR = 1;            while (dvd >= (sub << 1))            { sub <<= 1; subR <<= 1; }            dvd -= sub;            result += subR;        }        return sign * result;    }}

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