[一起来刷leetcode吧][24]--No.301 Remove Invalid Parentheses

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这是leetcode的第301题--Remove Invalid Parentheses

  题目 Remove the minimum number of invalid parentheses in order to make the input string valid. Return all possible results.

Note: The input string may contain letters other than the parentheses ( and ).

Examples:
"()())()" -> ["()()()", "(())()"]
"(a)())()" -> ["(a)()()", "(a())()"]
")(" -> [""]

  思路 代码是别人的，我写得太矬。我写了几十行代码，人家只用了几行:sob:。这道题可以用BFSFS，注意到对于一个坏括号列，要变成好括号列，要删去的括号种类（即(或者)）和数目是固定的，所以可以逐个搜索删去1，2，3...个后，是否有好括号列。不过这样不足的一点就是太慢了，搜索得太多了。   

show me the code

class Solution(object):
    def judgePoint24(self, nums):
        """
        :type nums: List[int]
        :rtype: bool
        """
        return self.exam(nums,4)
    def div(a,b):
        if abs(b)< 0.01:
            return 10000
        return a/float(b)
    from operator import mul,add,sub
    ops = [mul,add,sub,div]
    def exam(self,nums,n):
        if n is 1 :
            if abs(nums[0] -24)<0.01:
                return True
            else:return False
        for op in self.ops:
            for i in range(n-1):
                for j in range(i 1,n):
                    rst1 = op(nums[i],nums[j])
                    rst2 = op(nums[j],nums[i]) 
                    tmp = nums[:]
                    tmp.remove(nums[i])
                    tmp.remove(nums[j])
                    tmp.append(rst1)
                    if self.exam(tmp,n-1):
                        return True
                    if rst1 != rst2:
                        tmp[-1] = rst2
                        if self.exam(tmp,n-1):
                            return True
        return False