leetcode(44)-----227. Basic Calculator II

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Implement a basic calculator to evaluate a simple expression string.

The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero.

Example 1:

Input: "3+2*2"
Output: 7

Example 2:

Input: " 3/2 "
Output: 1

Example 3:

Input: " 3+5 / 2 "
Output: 5

Note:

• You may assume that the given expression is always valid.
• Do not use the eval built-in library function.
  

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思路分析：

• 1.现将字符串去空格，将字符串中转换成列表
• 2.将字符串中相连的数字 放在列表里 只占一位
• 3.计算整理好的数据

但是提交 显示超时了。。。。。。今天没时间了，回头再改改。

class Solution:
    def calculate(self, s):
        s = list(s.replace(' ', ''))       

        i = j = l = 0
        n = len(s)
        while (i < n):
            if s[j].isdigit():
                curnum = cum = 0
                while (i < n and s[i].isdigit()):
                    curnum = curnum * 10 + int(s[i])
                    i += 1
                    cum += 1
                s[j] = curnum
                l += 1
                for k in range(1, cum):
                    s.remove(s[2 * l - 1])
                    i -= 1
            j += 2
            n -= cum - 1
            i += 1
        i = 0

        while i < len(s) - 1:
            if s[i] == '*':
                s[i] = int(s[i - 1]) * int(s[i + 1])
                del s[i + 1]
                del s[i - 1]
                i = 0
            elif s[i] == '/':
                s[i] = int(s[i - 1]) // int(s[i + 1])
                del s[i + 1]
                del s[i - 1]
                i = 0
            i += 1
        i = 0
        while i < len(s) - 1:
            if s[i] == '+':
                s[i] = int(s[i - 1]) + int(s[i + 1])
                del s[i + 1]
                del s[i - 1]
                i = 0
            elif s[i] == '-':
                s[i] = int(s[i - 1]) - int(s[i + 1])
                del s[i + 1]
                del s[i - 1]
                i = 0
            i += 1
        return s[0]

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