[LeetCode] Fraction Addition and Subtraction 分数加减法

2023年10月31日 309次阅读 来源: Grandyang

 

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.

 

这道题让我们做分数的加减法，给了我们一个分数加减法式子的字符串，然我们算出结果，结果当然还是用分数表示了。那么其实这道题主要就是字符串的拆分处理，再加上一点中学的数学运算的知识就可以了。这里我们使用字符流处理类来做，每次按顺序读入一个数字，一个字符，和另一个数字。分别代表了分子，除号，分母。我们初始化分子为0，分母为1，这样就可以进行任何加减法了。中学数学告诉我们必须将分母变为同一个数，分子才能相加，为了简便，我们不求最小公倍数，而是直接乘上另一个数的分母，然后相加。不过得到的结果需要化简一下，我们求出分子分母的最大公约数，记得要取绝对值，然后分子分母分别除以这个最大公约数就是最后的结果了，参见代码如下：

 

class Solution {
public:
    string fractionAddition(string expression) {
        istringstream is(expression);
        int num = 0, dem = 0, A = 0, B = 1; 
        char c;
        while (is >> num >> c >> dem) {
            A = A * dem + num * B;
            B *= dem;
            int g = abs(gcd(A, B));
            A /= g;
            B /= g;
        }
        return to_string(A) + "/" + to_string(B);
    }
    int gcd(int a, int b) {
        return (b == 0) ? a : gcd(b, a % b);
    }
};

 

参考资料：

https://discuss.leetcode.com/topic/89991/concise-java-solution

https://discuss.leetcode.com/topic/90061/small-simple-c-java-python

 

    原文作者：Grandyang
    原文地址: http://www.cnblogs.com/grandyang/p/6954197.html
    本文转自网络文章，转载此文章仅为分享知识，如有侵权，请联系博主进行删除。
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