C++实现LeetCode(29.两数相除)

[LeetCode] 29. Divide Two Integers 两数相除

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;
    }
};

到此这篇关于C++实现LeetCode(29.两数相除)的文章就介绍到这了,更多相关C++实现两数相除内容请搜索我们以前的文章或继续浏览下面的相关文章希望大家以后多多支持我们！

赞 (0)

相关推荐

  • C++实现LeetCode(25.每k个一组翻转链表)

    [LeetCode] 25. Reverse Nodes in k-Group 每k个一组翻转链表 Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. k is a positive integer and is less than or equal to the length of the linked list. If the number of n

  • C++实现LeetCode(23.合并k个有序链表)

    [LeetCode] 23. Merge k Sorted Lists 合并k个有序链表 Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. Example: Input: [ 1->4->5, 1->3->4, 2->6 ] Output: 1->1->2->3->4->4->5->6 这

  • C++实现LeetCode(27.移除元素)

    [LeetCode] 27. Remove Element 移除元素 Given an array nums and a value val, remove all instances of that value in-place and return the new length. Do not allocate extra space for another array, you must do this by modifying the input array in-place with

  • C++实现LeetCode(28.实现strStr()函数)

    [LeetCode] 28. Implement strStr() 实现strStr()函数 Implement strStr(). Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. Example 1: Input: haystack = "hello", needle = "ll" Output: 2 E

  • C++实现LeetCode(26.有序数组中去除重复项)

    [LeetCode] 26. Remove Duplicates from Sorted Array 有序数组中去除重复项 Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length. Do not allocate extra space for another array, you must do this

  • C++实现LeetCode(24.成对交换节点)

    [LeetCode] 24. Swap Nodes in Pairs 成对交换节点 Given a linked list, swap every two adjacent nodes and return its head. You may not modify the values in the list's nodes, only nodes itself may be changed. Example: Given 1->2->3->4 , you should return t

  • C++实现LeetCode(83.移除有序链表中的重复项)

    [LeetCode] 83. Remove Duplicates from Sorted List 移除有序链表中的重复项 Given a sorted linked list, delete all duplicates such that each element appear only once. Example 1: Input: 1->1->2 Output: 1->2 Example 2: Input: 1->1->2->3->3 Output: 1-

  • C++实现LeetCode(29.两数相除)

    [LeetCode] 29. Divide Two Integers 两数相除 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

  • 在Python中获取两数相除的商和余数方法

    方法一:可以使用//求取两数相除的商.%求取两数相除的余数.[/在Python中获取的是相除的结果,一般为浮点数] 方法二:使用divmod()函数,获取商和余数组成的元祖 实例代码: #!/usr/bin/python3 # -*- coding: utf-8 -*- a = int(input(u"输入被除数: ")) b = int(input(u"输入除数:")) div = a // b mod = a % b print("{} / {} =

  • Java实现LeetCode（两数之和）

    给定一个整数数组和一个目标值,找出数组中和为目标值的两个数. 你可以假设每个输入只对应一种答案,且同样的元素不能被重复利用. 示例: 给定 nums = [2, 7, 11, 15], target = 9 因为 nums[0] + nums[1] = 2 + 7 = 9 所以返回[0, 1] 思路一:最直接的思维,两次遍历查询,时间复杂度O(N*N). 代码: public static int[] twoSum1(int[] nums, int target) { int[] label =

  • C++实现LeetCode(170.两数之和之三 - 数据结构设计)

    [LeetCode] 170. Two Sum III - Data structure design 两数之和之三 - 数据结构设计 Design and implement a TwoSum class. It should support the following operations: add and find. add - Add the number to an internal data structure. find - Find if there exists any pai

  • C++实现LeetCode(167.两数之和之二 - 输入数组有序)

    [LeetCode] 167.Two Sum II - Input array is sorted 两数之和之二 - 输入数组有序 Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number. The function twoSum should return indices of t

  • Java两整数相除向上取整的方式详解(Math.ceil())

    Java两整数相除向上取整的方式详解(Math.ceil())

    目录 前言: 方式一: 添加三目运算符逻辑代码 方式二:使用ceil函数 方式三:其他逻辑 最后总结 附:java向上取整函数Math.ceil() 前言: Java中两个整数相除,如果不能整除,默认是向下取整的.例如:11 除以 3 的结果是 3.然而,某些情况下(eg. 把11个糖果,每3个分一堆,不足三个也分成一堆,可以分几堆?),我们需要向上取整,这样的情况该如果处理呢? 方式一: 添加三目运算符逻辑代码 x / y + (x % y != 0 ? 1 : 0); 这种方法逻辑上很简单,

  • C++实现LeetCode（三数之和）

    Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero. Note: Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c) The solut

  • C++实现LeetCode(18.四数之和)

    [LeetCode] 18. 4Sum 四数之和 Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target. Note: Elements in a quadruplet (a,b,c,d) must be

  • C++实现LeetCode(202.快乐数)

    [LeetCode] 202.Happy Number 快乐数 Write an algorithm to determine if a number is "happy". A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits,

  • 详解JavaScript中任意两数加减的解决方案

    目录 写在前面 分析填坑思路 解决整数加减的坑 转换科学计算 解决整数减法的坑 解决小数加法的坑 解决小数减法的坑 解决整数加小数的通用问题 总结 写在前面 本文是从初步解决到最终解决的思路,文章篇幅较长 虽然是一篇从0开始的文章,中间的思维跳跃可能比较大 代码的解析都在文章的思路分析和注释里,全文会帮助理解的几个关键词 1.Number.MAX_SAFE_INTEGER 和 Number.MIN_SAFE_INTEGER 2.15长度的字符串 3.padStart 和 padEnd 分析填坑思

随机推荐