"-":"";9stringrem;10inti =0;1112num =abs(num);13den =abs(den);14res += to_string(num /den);15num = (num % den) *10;1617while(num) {18if(record.find(num) ==record.end())19record[num] =i;20else{21rem = rem.substr(0, record[num]) +"("+ rem.substr(record[num]) +...
--- 本期向大家介绍一些 Python 中用于处理数字和数学方面的标准库。...math 数学模块作为 LeetCode Python 环境中默认导入的标准库模块之一,math模块提供了很多非常有用的数字和数学方面的函数。...assert math.isclose(1.0, 1.0 + 1e-10) == True assert math.isclose(1.0, 1.0 + 1e8) == False asser...
Given two integers representing the numerator and denominator of a fraction, return the fraction in string format. If the fractional part is repeating, enclose the repeating part in parentheses. Example 1: Input: numerator = 1, denominator = 2 Output: "0.5" Example 2: Input: numerator = 2, ...
【leetcode】1290. Convert Binary Number in a Linked List to Integer 2019-12-23 09:54 −题目如下: Given head which is a reference node to a singly-linked list. The value of each node in the linked list is either 0 or 1. The lin... ...
【leetcode】1290. Convert Binary Number in a Linked List to Integer 2019-12-23 09:54 −题目如下: Given head which is a reference node to a singly-linked list. The value of each node in the linked list is either 0 or 1. The lin... ...
Fraction to Recurring Decimal Given two integers representing the numerator and denominator of a fraction, return the fraction in string format. If the fractional part is repeating, enclose the repeating part in parentheses. For example, Given numerator = 1, denominator = 2, return "0.5". ...
Output For each line of input, print the fraction, its decimal expansion through the first occurrence of the cycle to the right of the decimal or 50 decimal places (whichever comes first), and the length of the entire repeating cycle. ...
Special thanks to@Shangrilafor adding this problem and creating all test cases. Subscribeto see which companies asked this question 解法:分数一定可以表示为不循环小数(包括整数)或者无限循环小数。无限不循环小数即是无理数。设置一个哈希表,存储每一次的余数,以及该余数在返回结果result中的下标。每一次得到新...
leetcode166. Fraction to Recurring Decimal 题意:给出分子和分母,用字符串表示结果,循环的部分用“()”括起来。 思路: 模拟笔算, 最初根据1/99的情况,设想用path记录余数相同时的间隔步长,但只能用于处理(00..0n)的情况,其中n为[1,9]; 后来采取网上看到用map记录<余数,出现位置>的想法。
1. When numerator is 0, return "0". Check this corner case, because 0 / -5 will return -0. 2. Use long long int for divd and divs, mainly for divs. Because when divs = INT_MIN, fabs(divs) is large. 3. Do not forget to rest *= 10 to get the next level. ...