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题目描述：
Divide two integers without using multiplication, division and mod operator.
If it is overflow, return INT_MAX.
提示中谈到Binary Search，由此可想到解决方案。
solution：
int divide(int dividend, int divisor) { if(divisor == 0) return INT_MAX; if(dividend == INT_MIN && divisor == -1) return INT_MAX; int sign = (dividend > 0 ^ divisor > 0) ? -1 : 1; long long end = abs( (long long)dividend); long long sor = abs( (long long)divisor); long long res = 0; while (end >= sor) { long long tmp = sor; long long power = 1; while ( (tmp << 1) < end) { power <<= 1; tmp <<= 1; } res += power; end -= tmp; } return sign > 0 ? res : -res; }
参考资料：https://leetcode.com/discuss/11358/simple-o-log-n-2-c-solution
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原文地址：http://www.cnblogs.com/gattaca/p/4581825.html
