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  • Java [leetcode 29]Divide Two Integers

    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);
    	}
    }
    
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  • 原文地址:https://www.cnblogs.com/zihaowang/p/4507605.html
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