【LeetCode】29. Divide Two Integers（C++）

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

题目：

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: [ $−2^{31}$, $2^{31}− 1$]. For the purpose of this problem, assume that your function returns $2^{31}− 1$ when the division result overflows.

理解：

不能使用除号，一下就想到了循环。写完果不其然，超时了。
看了discuss，用了位运算。要比循环快多了。

实现：

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