29. Divide Two Integers

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

If it is overflow, return MAX_INT.

不使用乘法、除法和取余操作实现两数相除。
若数值溢出，返回MAX_INT.

#include "limits.h"#include <algorithm>int DivideTwoIntegers::divide(int dividend, int divisor){    //数值需要是long long 类型    //因为输入的都是int型，比如被除数是-2147483648，在int范围内，当除数是-1时，结果就超出了int范围，需要返回INT_MAX    if (divisor == 0 || (dividend == INT_MIN && divisor == -1)) return INT_MAX;    long long m = abs((long long)dividend), n = abs((long long)divisor), res = 0;    //从最大倍数开始减,实现O(logN)的时间复杂度    while (m >= n)    {        //t做为临时变量，将被左移扩展至不超过m的最大倍数        long long t = n, p = 1;        while (m >= (t << 1))        {            t <<= 1;            p <<= 1;        }        res += p;        m -= t;    }    //注意符号问题    if ((dividend < 0) ^ (divisor < 0)) res = -res;    return res;}