241. Different Ways to Add Parentheses——本质:DFS

Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are +, - and *.


Example 1

Input: "2-1-1".

((2-1)-1) = 0
(2-(1-1)) = 2

Output: [0, 2]


Example 2

Input: "2*3-4*5"

(2*(3-(4*5))) = -34
((2*3)-(4*5)) = -14
((2*(3-4))*5) = -10
(2*((3-4)*5)) = -10
(((2*3)-4)*5) = 10

Output: [-34, -14, -10, -10, 10]

复制代码
class Solution(object):
    def diffWaysToCompute(self, input):
        """
        :type input: str
        :rtype: List[int]
        A: (2*3)-4*5
        ((2*3)-4)*5=>1
        (2*3)-(4*5)=>2
        B:
        2*(3-4)*5
        (2*(3-4))*5=>1
        2*((3-4)*5)=>2
        """
        import re
        tokens = re.split('(\D)', input)
        nums = map(int, tokens[::2])
        ops = map({'+': operator.add, '-': operator.sub, '*': operator.mul}.get, tokens[1::2])
        def build(lo, hi):
            if lo == hi:
                return [nums[lo]]
            ans = []
            for i in range(lo, hi):
                for a in build(lo, i):
                    for b in build(i+1, hi):
                        ans.append(ops[i](a, b))    
            return ans
        return build(0, len(ops))        
复制代码

 

re.split("(\D)", "2+3-1")
['2', '+', '3', '-', '1']

posted @   bonelee  阅读(297)  评论(0编辑  收藏  举报
编辑推荐:
· 记一次.NET内存居高不下排查解决与启示
· 探究高空视频全景AR技术的实现原理
· 理解Rust引用及其生命周期标识(上)
· 浏览器原生「磁吸」效果!Anchor Positioning 锚点定位神器解析
· 没有源码,如何修改代码逻辑?
阅读排行:
· 全程不用写代码,我用AI程序员写了一个飞机大战
· MongoDB 8.0这个新功能碉堡了,比商业数据库还牛
· 记一次.NET内存居高不下排查解决与启示
· 白话解读 Dapr 1.15:你的「微服务管家」又秀新绝活了
· DeepSeek 开源周回顾「GitHub 热点速览」
点击右上角即可分享
微信分享提示