592. Fraction Addition and Subtraction

本文介绍了一种解决分数加减表达式计算的方法,通过使用数学算法,将给定的分数表达式转换为不可约形式的计算结果。文章提供了详细的Java实现代码,包括如何处理输入字符串、计算分子和分母、简化分数以及返回最终结果。

Given a string representing an expression of fraction addition and subtraction, you need to return the calculation result in string format. The final result should be irreducible fraction. If your final result is an integer, say 2, you need to change it to the format of fraction that has denominator 1. So in this case, 2 should be converted to 2/1.

 

Example 1:

Input:"-1/2+1/2"
Output: "0/1"

 

Example 2:

Input:"-1/2+1/2+1/3"
Output: "1/3"

 

Example 3:

Input:"1/3-1/2"
Output: "-1/6"

 

Example 4:

Input:"5/3+1/3"
Output: "2/1"

 

Note:

  1. The input string only contains '0' to '9''/''+' and '-'. So does the output.
  2. Each fraction (input and output) has format ±numerator/denominator. If the first input fraction or the output is positive, then '+'will be omitted.
  3. The input only contains valid irreducible fractions, where the numerator and denominator of each fraction will always be in the range [1,10]. If the denominator is 1, it means this fraction is actually an integer in a fraction format defined above.
  4. The number of given fractions will be in the range [1,10].
  5. The numerator and denominator of the final result are guaranteed to be valid and in the range of 32-bit int.

 

Approach #1: Math. [Java]

class Solution {
    public String fractionAddition(String expression) {
        Scanner sc = new Scanner(expression).useDelimiter("/|(?=[-+])");
        int A = 0, B = 1;
        while (sc.hasNext()) {
            int a = sc.nextInt(), b = sc.nextInt();
            A = A * b + a * B;
            B *= b;
            int g = GCD(A, B);
            A /= g;
            B /= g;
        }
        return A + "/" + B;
    }
    
    public int GCD(int a, int b) {
        return b == 0 ? Math.abs(a) : GCD(b, a%b);
    }
}

  

Reference:

https://leetcode.com/problems/fraction-addition-and-subtraction/discuss/103384/Small-simple-C%2B%2BJavaPython

 

转载于:https://www.cnblogs.com/ruruozhenhao/p/10864472.html

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值