力扣题-12.15

[力扣刷题攻略] Re：从零开始的力扣刷题生活

力扣题1：592. 分数加减运算

解题思想：首先通过+对表达式进行分离，然后利用分数的加法原则进行计算，最后除以最大公因数即可

class Solution(object):def fractionAddition(self, expression):""":type expression: str:rtype: str"""modified_expression = expression.replace('-', '+-')temp = modified_expression.split("+")up = 0down = 1for i in range(len(temp)):if temp[i] != '':temp_up = int(temp[i].split('/')[0])temp_down = int(temp[i].split('/')[1])up = up * temp_down + down * temp_updown = down * temp_downfor j in range(min(abs(up),abs(down)),1,-1):if down%j == 0 and up%j ==0:down = down/jup = up/jbreakif up == 0:return "0/1"else:return str(int(up)) + "/" + str(int(down))

class Solution {
public:string fractionAddition(string expression) {std::string modified_expression = replace_minus_with_plus_dash(expression);std::vector<std::string> temp = split_expression(modified_expression, '+');int up = 0;int down = 1;for (size_t i = 0; i < temp.size(); ++i) {if (!temp[i].empty()) {int temp_up = std::stoi(split_fraction(temp[i], '/')[0]);int temp_down = std::stoi(split_fraction(temp[i], '/')[1]);up = up * temp_down + down * temp_up;down = down * temp_down;}}for (int j = std::min(std::abs(up), std::abs(down)); j > 1; --j) {if (down % j == 0 && up % j == 0) {down = down / j;up = up / j;break;}}if (up == 0) {return "0/1";} else {return std::to_string(up) + "/" + std::to_string(down);}}private:std::string replace_minus_with_plus_dash(std::string str) {size_t found = str.find("-");while (found != std::string::npos) {str.replace(found, 1, "+-");found = str.find("-", found + 2);}return str;}std::vector<std::string> split_expression(std::string str, char delimiter) {std::vector<std::string> result;size_t pos = 0;while ((pos = str.find(delimiter)) != std::string::npos) {result.push_back(str.substr(0, pos));str.erase(0, pos + 1);}result.push_back(str);return result;}std::vector<std::string> split_fraction(std::string str, char delimiter) {std::vector<std::string> result;size_t pos = str.find(delimiter);result.push_back(str.substr(0, pos));result.push_back(str.substr(pos + 1));return result;}
};