zoukankan      html  css  js  c++  java
  • 29. Divide Two Integers

    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.

    只能用加减来实现除法-> recutsion。主要考虑以下几种情况:

    1. 溢出 -> 用long

    2. 被除数为0 或者 被除数小于除数 -> 返回0

    3. 符号问题

    time: O(logn), space: O(logn)

    class Solution {
        public int divide(int dividend, int divisor) {
            int sign = 1;
            if((dividend < 0 && divisor > 0) || (dividend > 0 && divisor < 0)) sign = -1;
            
            long longdividend = Math.abs((long)dividend);
            long longdivisor = Math.abs((long)divisor);
            if(longdividend == 0 || longdividend < longdivisor) return 0;
            
            long longres = helper(longdividend, longdivisor);
            
            int res = (int)longres;
            if(longres > Integer.MAX_VALUE) {
                res = sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
            }
            return sign * res;
        }
        public long helper(long dividend, long divisor) {
            if(dividend < divisor) return 0;
            long sum = divisor;
            long multiple = 1;
            while(sum + sum <= dividend) {
                sum += sum;
                multiple += multiple;
            }
            return multiple + helper(dividend - sum, divisor);
        }
    }
  • 相关阅读:
    1348:【例4-9】城市公交网建设问题
    1392:繁忙的都市(city)
    1381:城市路(Dijkstra)
    初识微积分
    进阶数论(1)逆元
    [题解] Codeforces Round #549 (Div. 2) B. Nirvana
    简单数论之整除&质因数分解&唯一分解定理
    [题解]ybt1365:FBI树(fbi)
    [题解]一本通1240:查找最接近的元素
    [题解]NOIP2018(普及组)T1标题统计(title)
  • 原文地址:https://www.cnblogs.com/fatttcat/p/10135208.html
Copyright © 2011-2022 走看看