Solve a given equation and return the value of x in the form of string "x=#value". The equation contains only '+', '-' operation, the variable x and its coefficient.
If there is no solution for the equation, return "No solution".
If there are infinite solutions for the equation, return "Infinite solutions".
If there is exactly one solution for the equation, we ensure that the value of x is an integer.
Example 1:
Input: "x+5-3+x=6+x-2" Output: "x=2"
Example 2:
Input: "x=x" Output: "Infinite solutions"
Example 3:
Input: "2x=x" Output: "x=0"
Example 4:
Input: "2x+3x-6x=x+2" Output: "x=-1"
Example 5:
Input: "x=x+2" Output: "No solution"
Related Topics:
Math
Similar Questions:
// OJ: https://leetcode.com/problems/solve-the-equation/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
private:
int read(string &equation, int pos, bool &isNum, int &num) {
if (equation[pos] == '=') ++pos;
int sign = 1, N = equation.size();
if (equation[pos] == '+' || equation[pos] == '-') {
if (equation[pos] == '-') sign = -1;
++pos;
}
if (pos < N && equation[pos] == 'x') {
num = 1;
} else {
num = 0;
while (pos < N && isdigit(equation[pos])) {
num = num * 10 + equation[pos++] - '0';
}
}
isNum = true;
if (pos < N && equation[pos] == 'x') {
isNum = false;
++pos;
}
num *= sign;
return pos;
}
public:
string solveEquation(string equation) {
int i = 0, N = equation.size(), space = equation.find_first_of("=");
int coef = 0, val = 0;
while (i < N) {
bool isNum;
int num;
i = read(equation, i, isNum, num);
int sign = i <= space ? 1 : -1;
if (isNum) val -= sign * num;
else coef += sign * num;
}
if (coef) return "x=" + to_string(val / coef);
return val ? "No solution" : "Infinite solutions";
}
};