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))