一、 问题描述
Leecode第二十九题，题目为：
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.
问题理解为：
给定两个整数，将其分为被除数和除数，在不使用乘法、除法和余运算符的情况下对两个整数进行除法。
除除数后返回商。
整数除法应该向零截断。
例1:
输入:被除数= 10，除数= 3
输出:3
例2:
输入:被除数= 7，除数= -3
输出:2
说明:
被除数和除数都是32位带符号整数。
除数永远不会是0。
假设我们处理的环境只能存储32位带符号整数范围内的整数:[- 231,231 - 1]。对于这个问题，假设函数在除法结果溢出时返回231 - 1。
二、算法思路
1、
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三、实现代码
class Solution {
public:int divide(int dividend, int divisor) {if (divisor == 0 || (dividend == INT_MIN && divisor == -1)) return INT_MAX;long long x = abs((long long)dividend), y = abs((long long)divisor), res = 0;int sign = ((dividend < 0) ^ (divisor < 0)) ? -1 : 1;if (y == 1) return sign == 1 ? x : -x;while (x >= y) {long long t = y, p = 1;while (x >= (t << 1)) {t <<= 1;p <<= 1;}res += p;x -= t;}return sign == 1 ? res : -res;}
};