Divide Two Integers

題目描述

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.
(給出兩個數——被除數和除數，用這兩個數做出發運算，不能用乘法、除法和求餘運算；函數返回除法運算後的商，除數不能爲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.

題目分析

不能用除法，而除法求餘本身就是計算被除數中含有多少個完整的除數，可以使用while循環，讓被除數dividend不斷地去減除數divisor，減到最後小於除數就不能再減了，在這個循環中循環次數就是最後的結果絕對值，最後的結果還要加上一個符號。

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)
    
};