[leetcode 29]divide two integers

1 题目

Divide two integers without using multiplication, division and mod operator.

If it is overflow, return MAX_INT.

 

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2 思路

 

以前没见过，没思路。于是看别人的实现：https://leetcode.com/discuss/11358/simple-o-log-n-2-c-solution

 

使用的是移位操作：

例如：19 / 3

3 <<=1  ->6,    6<19 继续移位

6 <<=1 ->12,   12<19 继续移位

12 <<=1 ->24, 24 > 19不移位了

结果+ 4

19 － 12 ＝ 7重复移位

3 <<=1 ->6, 6<7 继续移位

6 <<=1 ->12 , 12 > 6 ,不移位了

结果+ 2

7-6=1

再比较

3 <<=1 ->6 ,6>1,不移位了

 

所以最后结果是6。

详情边界处理以及过程见代码。

 

3 代码

    public int divide(int dividend, int divisor) {
        if (divisor == 1) {
            return dividend;
        }
        //overflow
        if (dividend == Integer.MIN_VALUE && divisor == -1) {
            return Integer.MAX_VALUE;
        }

        int sign = ((dividend > 0) ^ (divisor > 0)) ? -1 : 1;
        
        int answer = 0;
        long former = Math.abs((long)dividend);
        long latter = Math.abs((long)divisor);
        while(former >= latter){
            long n = latter;
            int m = 1;
            while((n<<1) < former){
                n <<=1;
                m <<=1;
            }
            answer += m;
            former = former - n;
        }
        
        return sign * answer;
    }