力扣题:数字与字符串间转换-12.16
2023-12-16 05:20:33
力扣题-12.16
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
class Solution(object):
def solveEquation(self, equation):
"""
:type equation: str
:rtype: str
"""
number = 0
factor = 0
temp = 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 = 0
temp_factor = 0
if left[i] =='x':
temp_factor = 1
elif left[i][-1] =='x':
if left[i][:-1] =='-':
temp_factor = -1
else:
temp_factor = int(left[i][:-1])
else:
temp_number = int(left[i])
factor = factor+temp_factor
number = number+temp_number
right = right.replace('-', '+-')
right = right.split("+")
for i in range(len(right)):
if right[i]!='':
temp_number = 0
temp_factor = 0
if right[i] =='x':
temp_factor = -1
elif right[i][-1] =='x':
if right[i][:-1] =='-':
temp_factor = 1
else:
temp_factor = int(right[i][:-1]) *-1
else:
temp_number = int(right[i])*-1
factor = factor+temp_factor
number = number+temp_number
if 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;
}
}
}
};
文章来源:https://blog.csdn.net/yumeng3866/article/details/134931935
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