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Question Solution
Divide two integers without using multiplication, division and mod operator.
If it is overflow, return MAX_INT.
这道题要不是参考别人的答案，恐怕较难得到思路。总体而言就是二分法。二分查找是很熟悉的内容了，但是真的将相关方法到应用的时候还是觉得略有困难。
参考文献：http://www.cnblogs.com/panda_lin/p/divide_two_integers.html
这道题算是一种很不错的经验吧。把代码奉上：
class Solution { public: unsigned long long newdivide(unsigned long long dividend, unsigned long long divisor) { unsigned long long temp=divisor,left; unsigned long long result=1; if (dividend < divisor) return 0; while (temp <= dividend) { left = dividend - temp; temp <<= 1; if (temp > dividend) break; else result <<= 1; } return result + newdivide(left, divisor); } int divide(int dividend, int divisor) { if (divisor == 0) throw 1;//抛出一个错误出去 bool positive = (dividend >= 0 && divisor > 0) || (dividend <= 0 && divisor < 0); unsigned long long _dividend = abs((long long)dividend); unsigned long long _divisor = abs((long long)divisor); long long tempresult=positive ? newdivide(_dividend, _divisor) : -1 * newdivide(_dividend, _divisor); if (abs(tempresult) >= INT_MAX) tempresult = positive ? INT_MAX : INT_MIN; return tempresult; } };
再学习一下二分法的内容：
HappyLeetcode45:Divide Two Integers
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原文地址：http://www.cnblogs.com/chengxuyuanxiaowang/p/4233458.html
