leetcode: 29. Divide Two Integers

最新推荐文章于 2022-05-31 11:11:52 发布

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最新推荐文章于 2022-05-31 11:11:52 发布

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本文介绍了一种在不使用乘法、除法和取模运算符的情况下，进行两整数相除的算法实现。该算法利用位操作和递归调用，通过二分查找的方式寻找商的值，实现了O(log(n))的时间复杂度和O(1)的空间复杂度。

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Difficulty

Medium

Description

Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator.

Return the quotient after dividing dividend by divisor.

The integer division should truncate toward zero.

Example 1:

Input: dividend = 10, divisor = 3
Output: 3
Example 2:

Input: dividend = 7, divisor = -3
Output: -2

Solution

Time Complexity: O(log(n))
Space Complexity: O(1)

class Solution {
    public int divide(int dividend, int divisor) {
        int sign = 1;
		if ((dividend > 0 && divisor < 0) || (dividend < 0 && divisor > 0))
			sign = -1;
		long ldividend = Math.abs((long) dividend);
		long ldivisor = Math.abs((long) divisor);

		if (ldividend == 0 || ldividend < ldivisor) return 0;
		
		int ans;
		long div = ldivide(ldividend, ldivisor);

		if (div > Integer.MAX_VALUE)
			ans = (sign == 1) ? Integer.MAX_VALUE : Integer.MIN_VALUE;
		else
			ans = (int) (sign * div);
		
		return ans;
    }

	public long ldivide(long dividend, long divisor) {
		if (dividend < divisor) return 0;
		
		long sum = divisor;
		long mul = 1;

		while ((sum + sum) <= dividend)   //二分查找
		{
			sum += sum;
			mul += mul;
		}
		return mul + ldivide(dividend - sum, divisor);
	}
}