leetcode---兩數相除

1.題目描述

    給定兩個32位的除數和被除數，不使用除法求出商數

2.問題求解

   除數設爲n,被除數設爲m,可以得到:m=n*(2^0+2^1+....2^t)，由於被除數和除數都是32位，所以t<=32,時間複雜度爲O(32)=O(1)

   注意數字越界

3.代碼實現

   

class Solution {
    public int divide(int dividend, int divisor) {
            boolean isNegative=((dividend<0&&divisor>0))||((dividend>0&&divisor<0));
            long extractDividend=Math.abs(((long)dividend));
            long extractDivisor=Math.abs(((long)divisor));
           
            long result=divideRecur(extractDividend,extractDivisor);
           
            result= isNegative?((0-result)):(result);
            if(result>Integer.MAX_VALUE)
            {
                return Integer.MAX_VALUE;
            }
            if(result<Integer.MIN_VALUE)
            {
                return Integer.MAX_VALUE;
            }
            return (int)result;
        
    }
    
    private long divideRecur(long dividend,long divisor)
    {
         long tmpDivisor=divisor;
        if(dividend<divisor)
        {
            return 0;
        }
        
        long result=1;
        while(divisor<=dividend)
        {
            divisor<<=1;
            result<<=1;
        }
        result>>=1;
        divisor>>=1;
        return result+divideRecur(dividend-divisor,tmpDivisor);
    }
    
    
}