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题目描述：
Divide two integers without using multiplication, division and mod operator.
If it is overflow, return MAX_INT.
解题思路：
把除数表示为：dividend = 2^i * divisor + 2^(i-1) * divisor + ... + 2^0 * divisor。这样一来，我们所求的商就是各系数之和了，而每个系数都可以通过移位操作获得。
详细解说请参考：http://blog.csdn.net/whuwangyi/article/details/40995863
代码如下：
public class Solution {
public int divide(int dividend, int divisor) {
boolean flag = (dividend > 0 && divisor > 0)
|| (dividend < 0 && divisor < 0);
long absDividend = Math.abs((long) dividend);
long absDivisor = Math.abs((long) divisor);
long quotient = dividePositive(absDividend, absDivisor);
if (flag && quotient > Integer.MAX_VALUE)
return Integer.MAX_VALUE;
return flag ? (int) quotient : -(int) quotient;
}
public long dividePositive(long dividend, long divisor) {
if (dividend < divisor)
return 0;
long quotient = 1;
long originalDivisor = divisor;
while (dividend >= (divisor << 1)) {
quotient <<= 1;
divisor <<= 1;
}
return quotient + dividePositive(dividend - divisor, originalDivisor);
}
}
Java [leetcode 29]Divide Two Integers
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原文地址：http://www.cnblogs.com/zihaowang/p/4507605.html
