Leetcode592. 分数加减

Leetcode592. Fraction Addition and Subtraction

题目

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.

解题分析

这道题本质上讲并不算很难，题目的要求很明显，就是让我们进行分数的加减法运算，需要注意的是，需要将分数化成最简形式。由于要将最后的结果化成最简形式，自然就考虑到了最大公约数的问题。
解题思路是非常清晰的，首先我们分别找到两个分数的分母和分子，然后根据分数加减的规则，将两个分数进行通分，最后将运算后的结果进行最简化。
这里在将字符串转化成数字的时候，用到了c++11新增的stoi函数，减少了一些操作；当然你也可以用原始的字符串转化成数字的方法代替。需要注意的是，在通分和化简的时候，通过最大公约数来进行相关的分数化简。

源代码

class Solution {public:    string fractionAddition(string expression) {        int n = 0, d = 1, index = 0, i, j;        if (expression[0] != '-') {            expression = "+" + expression;        }        while (index < expression.size()) {            for (i = index + 1; expression[i] != '/'; i++);            for (j = i + 1; expression[j] != '+' && expression[j] != '-' && j < expression.size(); j++);            int nn = stoi(expression.substr(index + 1, i - index - 1));            int dd = stoi(expression.substr(i + 1, j - i - 1));            int gcd1 = gcd(d, dd);            n = n * dd / gcd1 + (expression[index] == '-' ? -1 : 1) * nn * d / gcd1;            d = d * dd / gcd1;            index = j;        }        int gcd2 = gcd(abs(n), d);        return to_string(n / gcd2) + "/" + to_string(d / gcd2);    }    int gcd(int a, int b) {        while (b != 0) {            int temp = b;            b = a % b;            a = temp;        }        return a;    }};

以上是我对这道问题的一些想法，有问题还请在评论区讨论留言~