Divide Two Integers

题目描述

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.
(给出两个数——被除数和除数，用这两个数做出发运算，不能用乘法、除法和求余运算；函数返回除法运算后的商，除数不能为0)
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.

题目分析

不能用除法，而除法求余本身就是计算被除数中含有多少个完整的除数，可以使用while循环，让被除数dividend不断地去减除数divisor，减到最后小于除数就不能再减了，在这个循环中循环次数就是最后的结果绝对值，最后的结果还要加上一个符号。

程序示例

/**
 * @param {number} dividend
 * @param {number} divisor
 * @return {number}
 */
var divide = function(dividend, divisor) {
    var sig = true
    if(dividend>=0&&divisor>0 || dividend<=0&&divisor<0){
        sig = true
    }else{
        sig = false
    }
    if(dividend === 0){
        return 0
    }
    if(dividend==-2147483648 && divisor==-1){
        return 2147483647
    }
    var abs_num = 0
    var abs_did = Math.abs(dividend)
    var abs_div = Math.abs(divisor)
    if(abs_div==1){
        return parseInt((sig? '':'-') +abs_did)
    }
    while(abs_did){
        abs_did -= abs_div
        if(abs_did>=0){
            abs_num += 1
        }else{
            break
        }
    }
    return parseInt((sig? '':'-') +abs_num)
    
};