zoukankan      html  css  js  c++  java
  • Divide Two Integers leetcode

     题目:Divide Two Integers

    Divide two integers without using multiplication, division and mod operator.

    If it is overflow, return MAX_INT.

    看讨论区大神的思路:

    In this problem, we are asked to divide two integers. However, we are not allowed to use division, multiplication and mod operations. So, what else can we use? Yeah, bit manipulations.

    Let's do an example and see how bit manipulations work.

    Suppose we want to divide 15 by 3, so 15 is dividend and 3 is divisor. Well, division simply requires us to find how many times we can subtract the divisor from the the dividend without making the dividend negative.

    Let's get started. We subtract 3 from 15 and we get 12, which is positive. Let's try to subtract more. Well, we shift 3 to the left by 1 bit and we get 6. Subtracting 6 from 15 still gives a positive result. Well, we shift again and get 12. We subtract 12 from 15 and it is still positive. We shift again, obtaining 24 and we know we can at most subtract 12. Well, since 12 is obtained by shifting 3 to left twice, we know it is 4 times of 3. How do we obtain this 4? Well, we start from 1 and shift it to left twice at the same time. We add 4 to an answer (initialized to be0). In fact, the above process is like 15 = 3 * 4 + 3. We now get part of the quotient (4), with a remainder 3.

    Then we repeat the above process again. We subtract divisor = 3 from the remaining dividend = 3 and obtain 0. We know we are done. No shift happens, so we simply add 1 << 0 to the answer.

    Now we have the full algorithm to perform division.

    According to the problem statement, we need to handle some exceptions, such as overflow.

    Well, two cases may cause overflow:

    1. divisor = 0;
    2. dividend = INT_MIN and divisor = -1 (because abs(INT_MIN) = INT_MAX + 1).

    Of course, we also need to take the sign into considerations, which is relatively easy.

    Putting all these together, we have the following code.

    class Solution {
    public:
        int divide(int dividend, int divisor) {
            if (!divisor || (dividend == INT_MIN && divisor == -1))
                return INT_MAX;
            int sign = ((dividend < 0) ^ (divisor < 0)) ? -1 : 1;
            long long dvd = labs(dividend);
            long long dvs = labs(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; 
        }
    };

     1 #include<iostream>
     2 #include<limits>
     3 using namespace std;
     5 class Solution {
     6 public:
     7     int divide(int dividend, int divisor) 
     8     {
     9         int sign = ((dividend > 0) ^ (divisor > 0) ? -1 : 1);
    10         if (!divisor || (dividend==INT_MIN&&divisor==-1))
    11             return INT_MAX;
    12         long long divid = labs(dividend), divis = labs(divisor);
    13         long long  res = 0;
    14         while (divid >= divis)
    15         {
    16             long long temp = divis,multi_time=1;
    17             while (divid >= (temp<<1))
    18             {
    19                 temp <<= 1;
    20                 multi_time <<=1;
    21             }
    22             divid -= temp;
    23             res += multi_time;
    24         }
    25         return sign == 1 ? res:-res;
    26     }
    27 };
    28 int main()
    29 {
    30     Solution test;
    31     int res = test.divide(0, 1);
    32     cout << res << endl;
    33     return 0;
    34 }
    手里拿着一把锤子,看什么都像钉子,编程界的锤子应该就是算法了吧!
  • 相关阅读:
    一些C++11语言新特性
    项目管理计划应该包括哪些内容
    真相令人震惊!为什么越有钱的人,欠的钱越多?
    80后小伙返乡创业种植中药材,带领乡亲们脱贫致富
    Tableau
    知识点汇总
    决策树分析、EMV(期望货币值)
    信息系统项目管理师60天冲刺复习计划,2019下半年高项冲刺计划
    【系统分析师之路】系统分析师备考计划
    有一种规律:“劣币驱逐良币”,“坏人淘汰好人”(深度)
  • 原文地址:https://www.cnblogs.com/chess/p/5065378.html
Copyright © 2011-2022 走看看