[leetcode] 640. Solve the Equation

[leetcode] 640. Solve the Equation

Description

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"

分析

题目的意思是：给定一个字符串等式，求解x。

等式左边sign=1，等式右边sign=-1；b用来累加数，a用来累加符号x上的数。然后-b/a就是结果了哈，注意无解和无穷解的情况。无解a=0&&b!=0,无穷解a=0&&b=0。

代码

class Solution {public: string solveEquation(string equation) { int n=equation.size(); int j=0; int sign=1; int b=0,a=0; for(int i=0;ij){ b+=stoi(equation.substr(j,i-j))*sign; } j=i; }else if(equation[i]=='x'){ if(i==j||equation[i-1]=='+'){ a+=sign; }else if(equation[i-1]=='-'){ a-=sign; }else{ a+=stoi(equation.substr(j,i-j))*sign; } j=i+1; }else if(equation[i]=='='){ if(i>j){ b+=stoi(equation.substr(j,i-j))*sign; } sign=-1; j=i+1; } } if(j