[LeetCode] 29. Divide Two Integers ☆☆

 

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

If it is overflow, return MAX_INT.

 

解法：

　　这道题让我们求两数相除，而且规定我们不能用乘法，除法和取余操作。

　　采用位运算中的移位运算，左移一位相当于乘2，右移一位相当于除以2。假设求 a / b，将b左移n位后大于a，则结果 res += 1 << (n - 1)，将a更新 (a -= b << (n - 1)) 后进行同样操作，直到 a < b。

public class Solution {
    public int divide(int dividend, int divisor) {
        if (divisor == 0) {
            return Integer.MAX_VALUE;
        }
        
        if (dividend == Integer.MIN_VALUE && divisor == -1) {
            return Integer.MAX_VALUE;
        }
        
        boolean isNeg = (dividend > 0 && divisor < 0) || (dividend < 0 && divisor > 0);
        
        long left = Math.abs((long)dividend);
        long right = Math.abs((long)divisor);
        int result = 0;
        while (left >= right) {
            int times = 1;
            while ((right << times) <= left) {
                times++;
            }
            left -= (right << (times - 1));
            result += (1 << (times - 1));
        }
        
        return isNeg ? -result : result;
    }
}