python 回溯法模板详解

class Solution(object):

  #rtlist用来存储所有的返回所有排列，templist用来生成每个排列
  def backtrack(self,rtlist,templist,nums):
    if(len(templist) == len(nums)):
      rtlist.append(templist[:])
    else:
      for i in nums:

        if(i in templist): #如果在当前排列中已经有i了，就continue，相当于分支限界，即不对当前节点子树搜寻了
          continue
        templist.append(i)
        self.backtrack(rtlist,templist,nums)
        templist.pop() #把结尾的元素用nums中的下一个值替换掉，遍历下一颗子树

  def permute(self,nums):
    rtlist = []
    templist = []
    self.backtrack(rtlist,templist,nums)
    return rtlist

from collections import Counter

class Solution(object):

  def backtrack(self, rtlist, tmplist, counter, nums, length):
    if len(tmplist) == length:#回溯点
      rtlist.append(tmplist[:])
    else:
      for i in nums:#横向遍历
        if counter[i] == 0:#分支限界
          continue

        counter[i] -= 1
        tmplist.append(i)
        self.backtrack(rtlist, tmplist, counter, nums, length)#纵向遍历
        counter[i] += 1
        tmplist.pop()

  def permuteUnique(self, nums):

    rtlist, tmplist, counter = [], [], Counter(nums)
    length = len(nums)
    self.backtrack(rtlist, tmplist, counter, list(set(nums)), length)

    return rtlist

from collections import Counter

class Solution(object):

  def backtrack(self, rtlist, tmplist, level, counter, nums):
    if len(tmplist) == len(nums):
      rtlist.append(tmplist[:])
    else:
      for i in nums:
        if i in level or counter[i] == 0:
          continue

        counter[i] -= 1

        tmplist.append(i)
        level.add(i)
        self.backtrack(rtlist, tmplist, set(), counter, nums)

        counter[i] += 1
        tmplist.pop()


  def permuteUnique(self, nums):
    if not nums:
      return []
    rtlist, tmplist, level, counter = [], [], set(), Counter(nums)
    self.backtrack(rtlist, tmplist, level, counter, nums)

    return rtlist