leetcode 592. Fraction Addition and Subtraction 分数加法+C++stringstream的一个很好的示例

た入场券 2022-06-03 04:45 153阅读 0赞

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:  
The input string only contains ‘0’ to ‘9’, ‘/’, ‘+’ and ‘-‘. So does the output.  
Each fraction (input and output) has format ±numerator/denominator. If the first input fraction or the output is positive, then ‘+’ will be omitted.  
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.  
The number of given fractions will be in the range \[1,10\].  
The numerator and denominator of the final result are guaranteed to be valid and in the range of 32-bit int.

本题的题意很简单，就是针对分式做计算，最后返回最简分式作为结果

本题的关键件就是使用C++的stringstream来分割字符串，然后求最大公约数

建议和[leetcode 539. Minimum Time Difference C++中的stringstream真的很好用][leetcode 539. Minimum Time Difference C_stringstream] 一起学习

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    
    using namespace std;
    
    
    class Solution 
    {
    public:
        string fractionAddition(string expression)
        {
            stringstream ss(expression);
            int fenzi = 0, fenmu = 1, a, b;
            char c;
            while (ss >> a >> c >> b)
            {
                fenzi = fenzi*b + fenmu*a;
                fenmu = fenmu*b;
                int g = gcd(abs(fenzi), abs(fenmu));
                fenzi /= g;
                fenmu /= g;
            }
            return to_string(fenzi) + "/" + to_string(fenmu);
        }
    
        int gcd(int a, int b)
        {
            return b == 0 ? a : gcd(b, a%b);
        }
    };

[leetcode 539. Minimum Time Difference C_stringstream]: http://blog.csdn.net/jackzhang_123/article/details/78825210