29.Divide Two Integers

Tags: Medium Math Bit Manipulation

Links: https://leetcode.com/problems/divide-two-integers/

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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.

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class Solution {
public:
    int divide(int dividend, int divisor) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        long m = labs(dividend), n = labs(divisor);
        if (m < n) return 0;

        long tmp = n, p = 1, res = 0;
        while (m > (tmp << 1)) {
            tmp <<= 1;
            p <<= 1;
        }
        res += p + divide(m - tmp, n);

        if ((dividend < 0) ^ (divisor < 0)) res = -res;
        return res > INT_MAX ? INT_MAX : res;
    }
};

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Divide Two Integers.
Memory Usage: 8.4 MB, less than 14.00% of C++ online submissions for Divide Two Integers.

循环的部分的思路其实就是在模拟竖式除法。判断符号通过异或运算得出。

之所以需要判断是否大于INT_MAX，是因为存在INT_MIN / (-1)的情况。