leetcode 22.(括号生成）python（回溯4+剪枝）

通过分析生成的括号全排列，要求在排列生成的过程中右括号的数量小于等于左括号的数量。

基于上述想法，我们在回溯的过程中设置两个指针left和right；

代码如下：

class Solution:
    def generateParenthesis(self, n):

    	ans = []
    	
    	def back(tmp,left,right):

    		if len(tmp) == 2 * n:
    			ans.append(tmp)
    			return

    		if left < n:

    			back(tmp + '(', left + 1, right)

    		if right < left:

    			back(tmp + ')', left, right + 1)

    	back('',left = 0,right = 0)
    	return ans

下面是时间复杂度较高的：回溯的全排列+去重（剪枝）

class Solution:
    def generateParenthesis(self, n):
    	#首先生成数组
    	nums = '(' * n + ')' * n
    	# print(nums)
    	# for i in range(len(nums)):
    	# 	print(nums[i])
    	result = []
    	count1 = 0
    	count2 = 0
    	size = len(nums)
    	used = [False] * n * 2

    	def back(nums,tmp):


    		if len(tmp) == size:
    			result.append(tmp)
    			# print(result)
    			# count1 = 0
    			# count2 = 0

    			return#返回时候说明执行完毕


    		for i in range(len(nums)):
    			# print(used[i])

    			if not used[i]:
    				# print("####")
    				# print(i)
    				# print("*****")
    				# print(nums[i])

    				if i > 0 and nums[i] == nums[i - 1] and not used[i - 1]:
    					continue


    				count11 = (tmp + nums[i]).count('(')
    				count22 = (tmp + nums[i]).count(')')
    				# print(count11)
    				# print(count22)

    				if count11 < count22:
    					continue

    				used[i] = True


    				back(nums,tmp + nums[i])#
    				used[i] = False



    	back(nums,'')
    	return result