两数相除结果叫什么_两数相除余数最小是六

两数相除结果叫什么_两数相除余数最小是六Giventwointegers dividend and divisor,dividetwointegerswithoutusingmultiplication,divisionandmodoperator.Returnthequotie

大家好,欢迎来到IT知识分享网。

 

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.

 

这道题让我们求两数相除,而且规定不能用乘法,除法和取余操作,那么这里可以用另一神器位操作 Bit Manipulation,思路是,如果被除数大于或等于除数,则进行如下循环,定义变量t等于除数,定义计数p,当t的两倍小于等于被除数时,进行如下循环,t扩大一倍,p扩大一倍,然后更新 res 和m。这道题的 OJ 给的一些 test case 非常的讨厌,因为输入的都是 int 型,比如被除数是 -2147483648,在 int 范围内,当除数是  -1 时,结果就超出了 int 范围,需要返回 INT_MAX,所以对于这种情况就在开始用 if 判定,将其和除数为0的情况放一起判定,返回 INT_MAX。然后还要根据被除数和除数的正负来确定返回值的正负,这里采用长整型 long 来完成所有的计算,最后返回值乘以符号即可,代码如下:

 

解法一:

class Solution {
public:
    int divide(int dividend, int divisor) {
        if (dividend == INT_MIN && divisor == -1) return INT_MAX;
        long m = labs(dividend), n = labs(divisor), res = 0;
        int sign = ((dividend < 0) ^ (divisor < 0)) ? -1 : 1;
        if (n == 1) return sign == 1 ? m : -m;
        while (m >= n) {
            long t = n, p = 1;
            while (m >= (t << 1)) {
                t <<= 1;
                p <<= 1;
            }
            res += p;
            m -= t;
        }
        return sign == 1 ? res : -res;
    }
};

 

我们可以通过递归的方法来解使上面的解法变得更加简洁:

 

解法二:

class Solution {
public:
    int divide(int dividend, int divisor) {
        long m = labs(dividend), n = labs(divisor), res = 0;
        if (m < n) return 0;
        long t = n, p = 1;
        while (m > (t << 1)) {
            t <<= 1;
            p <<= 1;
        }
        res += p + divide(m - t, n);
        if ((dividend < 0) ^ (divisor < 0)) res = -res;
        return res > INT_MAX ? INT_MAX : res;
    }
};

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/29

 

参考资料:

https://leetcode.com/problems/divide-two-integers/

https://leetcode.com/problems/divide-two-integers/discuss/13524/summary-of-3-c-solutions

https://leetcode.com/problems/divide-two-integers/discuss/13407/C%2B%2B-bit-manipulations

https://leetcode.com/problems/divide-two-integers/discuss/142849/C%2B%2BJavaPython-Should-Not-Use-%22long%22-Int

 

LeetCode All in One 题目讲解汇总(持续更新中…)

免责声明:本站所有文章内容,图片,视频等均是来源于用户投稿和互联网及文摘转载整编而成,不代表本站观点,不承担相关法律责任。其著作权各归其原作者或其出版社所有。如发现本站有涉嫌抄袭侵权/违法违规的内容,侵犯到您的权益,请在线联系站长,一经查实,本站将立刻删除。 本文来自网络,若有侵权,请联系删除,如若转载,请注明出处:https://yundeesoft.com/29079.html

(0)

相关推荐

发表回复

您的邮箱地址不会被公开。 必填项已用 * 标注

关注微信