我想做个字符串解析的 用于四则运算各位有没有什么思路

百度找到一个，大家看看：
import math
import random
PRECISION = 1E-6
COUNT_OF_NUMBER = 4
NUMBER_TO_BE_CAL = 24
tmp_num1 = random.randint(1,13)
tmp_num2 = random.randint(1,13)
tmp_num3 = random.randint(1,13)
tmp_num4 = random.randint(1,13)
g_number = [tmp_num1, tmp_num2, tmp_num3, tmp_num4]
g_expression = ['', '', '', '']
for i in range(0, 4):
g_expression[i] = "%d" % g_number[i]
print(g_expression)
def solve(n):
if(1 == n):
if(math.fabs(NUMBER_TO_BE_CAL - g_number[0]) < PRECISION):
print(g_expression[0])
return True
else:
return False
else:
for i in range(0, n):
for j in range(i+1, n):
a = g_number[i]
b = g_number[j]
#**********************************
# 将剩下的有效数字往前挪，
# 由于两数计算结果保存在number[i]中，
# 所以将数组末元素覆盖number[j]即可
# *******************************
g_number[j] = g_number[n - 1]
expa = g_expression[i]
expb = g_expression[j]
g_expression[j] = g_expression[n - 1]
# 计算a+b
g_expression[i] = '(' + expa + '+' + expb + ')'
g_number[i] = a + b
if ( solve(n - 1) ) :
return True;
# 计算a-b
g_expression[i] = '(' + expa + '-' + expb + ')'
g_number[i] = a - b
if ( solve(n - 1) ) :
return True
# 计算b-a
g_expression[i] = '(' + expb + '-' + expa + ')'
g_number[i] = b - a
if ( solve(n - 1) ):
return True
# 计算(a*b)
g_expression[i] = '(' + expa + '*' + expb + ')'
g_number[i] = a * b
if ( solve(n - 1) ):
return True;
# 计算(a/b)
if (b != 0) :
g_expression[i] = '(' + expa + '/' + expb + ')'
g_number[i] = a / b
if ( solve(n - 1) ) :
return True
# 计算(b/a)
if (a != 0) :
g_expression[i] = '(' + expb + '/' + expa + ')'
g_number[i] = b / a
if ( solve(n - 1) ):
return True
# 恢复现场
g_number[i] = a
g_number[j] = b
g_expression[i] = expa
g_expression[j] = expb
return False
if(not solve(COUNT_OF_NUMBER)):
print('no solution')

难道就没有人学过编译原理吗