力扣题:数字与字符串间转换-12.16

力扣题-12.16

[力扣刷题攻略] Re：从零开始的力扣刷题生活

力扣题1：640. 求解方程

解题思想：首先将方程按照“=”进行划分，然后分别记录x的因数和常数项，最后进行返回的判断即可

在这里插入图片描述

class Solution(object):def solveEquation(self, equation):""":type equation: str:rtype: str"""number = 0factor = 0temp = equation.split("=")left = temp[0]right = temp[1]left = left.replace('-', '+-')left = left.split("+")print(left)for i in range(len(left)):if left[i]!='':temp_number = 0temp_factor = 0if left[i] =='x':temp_factor = 1elif left[i][-1] =='x':if left[i][:-1] =='-':temp_factor = -1else:temp_factor = int(left[i][:-1]) else:temp_number = int(left[i])factor = factor+temp_factornumber = number+temp_numberright = right.replace('-', '+-')right = right.split("+")for i in range(len(right)):if right[i]!='':temp_number = 0temp_factor = 0if right[i] =='x':temp_factor = -1elif right[i][-1] =='x':if right[i][:-1] =='-':temp_factor = 1else:temp_factor = int(right[i][:-1]) *-1else:temp_number = int(right[i])*-1factor = factor+temp_factornumber = number+temp_numberif factor ==0 and number==0:return "Infinite solutions"elif factor ==0:return "No solution"else:return "x="+str(number/factor*-1)

class Solution {
public:string solveEquation(string equation) {int number = 0;int factor = 0;std::vector<std::string> temp = splitEquation(equation, '=');std::string left = temp[0];std::string right = temp[1];left = replace_minus_with_plus_dash(left);std::vector<std::string> leftTokens = splitExpression(left, '+');processTokens(leftTokens, number, factor);right = replace_minus_with_plus_dash(right);std::vector<std::string> rightTokens = splitExpression(right, '+');processTokens(rightTokens, number, factor, true);if (factor == 0 && number == 0) {return "Infinite solutions";} else if (factor == 0) {return "No solution";} else {return "x=" + std::to_string(-number / factor);}}private:std::vector<std::string> splitEquation(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;}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> splitExpression(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;}void processTokens(const std::vector<std::string>& tokens, int& number, int& factor, bool isRight = false) {int multiplier = isRight ? -1 : 1;for (size_t i = 0; i < tokens.size(); ++i) {if (!tokens[i].empty()) {int temp_number = 0;int temp_factor = 0;if (tokens[i] == "x") {temp_factor = 1 * multiplier;} else if(tokens[i] == "-x"){temp_factor = -1 * multiplier;}else if (tokens[i].back() == 'x') {temp_factor = std::stoi(tokens[i].substr(0, tokens[i].size() - 1)) * multiplier;} else {temp_number = std::stoi(tokens[i]) * multiplier;}factor += temp_factor;number += temp_number;}}}
};