题目如下：
解题思路：本题的思路就是解析字符串，然后是小学时候学的解方程的思想，以"2x+3x-6x+1=x+2"，先把左右两边的x项和非x项进行合并，得到"-x+1=x+2"，接下来就是移项，把x项移到左边，常数项移到右边，得到"2x=-1",最后的解就是x=-1/2。对于任意一个表达式ax+b = cx+d来说，最终都能得到解x=(d-b)/(a-c)，这里要对(a-c)是否为0做判断，同时根据(d-b)是否为0等到Infinite solutions，No solution，常规解三种结果。
代码如下：
class Solution(object): def parse(self,expression): x,v = 0,0 if expression[0] == ‘-‘: expression = ‘0‘ + expression item = ‘‘ operators = [‘+‘, ‘-‘] operator = ‘‘ for i in (expression + ‘#‘): if i in operators or i == ‘#‘: if item == ‘x‘: item = ‘1x‘ if operator == ‘‘: operator = i if item[-1] == ‘x‘: x = int(item[:-1]) else: v = int(item) item = ‘‘ else: if operator == ‘+‘ and item[-1] == ‘x‘: x += int(item[:-1]) elif operator == ‘-‘ and item[-1] == ‘x‘: x -= int(item[:-1]) elif operator == ‘+‘ and item[-1] != ‘x‘: v += int(item) else: v -= int(item) item = ‘‘ operator = i else: item += i return x,v def solveEquation(self, equation): """ :type equation: str :rtype: str """ left,right = equation.split(‘=‘) lx,lv = self.parse(left) rx,rv = self.parse(right) if lx - rx == 0 and rv - lv == 0: return "Infinite solutions" elif lx - rx == 0 and rv - lv != 0: return "No solution" else: return "x=" + str((rv - lv)/(lx - rx))
【leetcode】640. Solve the Equation
原文：https://www.cnblogs.com/seyjs/p/9660467.html