leetcode -- Divide Two Integers

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

1.naive的做法是用除数一个一个减被除数，直到被除数<=0, 此方法会超时！！！

2.将被除数向左移位(x2),然后与被除数做比较，直到发现仅次于被除数的那个值，减去该值后继续(line:24-30)。

public class Solution {
    public static int divide(int dividend, int divisor) {
        if (dividend == 0 || divisor == 1)
            return dividend;
            
        int sign = 1;
        if (dividend < 0)
            sign = sign * -1;
        if (divisor < 0)
            sign = sign * -1;

        long divid = Math.abs((long)dividend);
        long divis = Math.abs((long)divisor);

        long[] a = new long[32];
        int i = 0;
        while (divis > 0 && divis <= divid) {
            a[i] = divis;
            divis = divis << 1;
            i++;
        }
        i--;
        long res = 0;
        while ( divid > 0 && i >= 0) {
            if(divid >= a[i]){
                divid = divid - a[i];
                res = res + (1 << i);
            }
            i--;
        }
        return (int)(res * sign) ;
    }
}

使用int时会出现整数溢出问题，INT_MIN(-2147483648)，比如 abs(-2147483648) = -2147483648，这里在12、13行全部使用long

另外+的优先级要比<<高

ref:http://fisherlei.blogspot.com/2012/12/leetcode-divide-two-integers.html