29. Divide Two Integers

Description

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

If it is overflow, return MAX_INT.

Solution

Binary search, time O(logn), space O(1)

比较蛋疼的题...起初想用减法做，但O(n)的时间会TLE，改成用加法做，利用类似二分的方式加速。

class Solution {
    public int divide(int dividend, int divisor) {
        if (divisor == 0) {
            return Integer.MAX_VALUE;
        }
        
        long x = dividend;
        long y = divisor;
        int sign = (x < 0 && y > 0) || (x > 0 && y < 0) ? -1 : 1;
        x = Math.abs(x);
        y = Math.abs(y);
        
        long z = divide(x, y);
        if (sign > 0 && z > Integer.MAX_VALUE) {    // overflow
            return Integer.MAX_VALUE;
        }
        
        return (int) (sign * z);
    }
    
    public long divide(long x, long y) {
        if (x < y) {
            return 0;
        }
        
        long sum = y;
        long multiple = 1;
        
        // Find the largest multiple so that (divisor * multiple <= dividend), 
        // moving with stride 1, 2, 4, 8, 16...2^n for performance reason
        while ((sum << 1) <= x) {
            sum <<= 1;
            multiple <<= 1;
        }
        
        return multiple + divide(x - sum, y);
    }
}