leetcode 282. 给表达式添加运算符

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给定一个仅包含数字 0-9 的字符串和一个目标值，在数字之间添加二元运算符（不是一元）+、- 或 * ，返回所有能够得到目标值的表达式。

示例 1:

输入: num = "123", target = 6
输出: ["1+2+3", "1*2*3"]

示例 2:

输入: num = "232", target = 8
输出: ["2*3+2", "2+3*2"]

示例 3:

输入: num = "105", target = 5
输出: ["1*0+5","10-5"]
示例 4: 
输入: num = "00", target = 0
输出: ["0+0", "0-0", "0*0"]
示例 5: 
输入: num = "3456237490", target = 9191
输出: []

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class Solution(object):
    def addOperators(self, num, target):
        def dfs(remain, curr_str, curr, prev):
            if not remain and curr == target:
                res.append(curr_str)
                
            for i in range(1, len(remain) + 1):
                if len(curr_str) == 0:   #avoid generate str begin with +-*
                    if not (i > 1 and remain[0] == '0'):   # avoid '0X' case be counted
                        dfs(remain[i:], remain[:i], int(remain[:i]), int(remain[:i]))
                else:
                    if not (i > 1 and remain[0] == '0'):
                        dfs(remain[i:], curr_str + '+' + remain[:i], curr + int(remain[:i]), int(remain[:i]))
                        dfs(remain[i:], curr_str + '-' + remain[:i], curr - int(remain[:i]), -int(remain[:i]))
                        # need take extra care for '*' case, a+b*c = a+b-b+b*c
                        dfs(remain[i:], curr_str + '*' + remain[:i], curr - prev + prev * int(remain[:i]), prev * int(remain[:i]))
            
        res = []
        dfs(num, '', 0, 0)
        return res
print(Solution().addOperators('105',5))