LeetCode第29题——Divide Two Integer

最新推荐文章于 2022-03-26 19:27:30 发布

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最新推荐文章于 2022-03-26 19:27:30 发布

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分类专栏： LeetCode刷题记录文章标签： leetcode 位运算

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本文介绍了一种在不使用乘法、除法和求余运算的情况下实现两个整数相除的方法。通过使用位运算和减法，算法能够在限定的时间内完成整数除法，并正确处理符号问题。C++代码示例展示了如何实现这一算法。

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问题描述
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
Note:

Both dividend and divisor will be 32-bit signed integers.
The divisor will never be 0.
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 231 − 1 when the division result overflows.
解法思路
不能使用乘、除和求余运算，理所当然想到用减法。但单纯使用减法会超时，需要使用位运算

C++代码

class Solution {
public:
    int divide(int dividend, int divisor) {
        if (dividend == INT_MIN && divisor == -1) {
            return INT_MAX;
        }
        long dvd = labs(dividend), dvs = labs(divisor), ans = 0;
        int sign = dividend > 0 ^ divisor > 0 ? -1 : 1;
        while (dvd >= dvs) {
            long temp = dvs, m = 1;
            while (temp << 1 <= dvd) {
                temp <<= 1;
                m <<= 1;
            }
            dvd -= temp;
            ans += m;
        }
        return sign * ans;
    }
};

算法结果
Runtime: 12 ms, faster than 83.88% of C++ online submissions for Divide Two Integers.
Memory Usage: 8 MB, less than 100.00% of C++ online submissions for Divide Two Integers.