Backtracking 编程范式python_backtracking模板python-优快云博客

本文链接：https://blog.youkuaiyun.com/BobChill/article/details/85094409

本文介绍了回溯编程的基本思想，包括选择、探索和撤销的范式，并通过两个实例展示了如何使用回溯法解决排列问题和掷骰子求组合问题。还提及了可以通过剪枝优化来提高算法效率。

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choose-> explore-> unchoose paradigm is the key! below is the general pseudo code

来看看例子一：Given a collection of distinct integers, return all possible permutations.

Input: [1,2,3]
Output:
[
  [1,2,3],
  [1,3,2],
  [2,1,3],
  [2,3,1],
  [3,1,2],
  [3,2,1]
]

class Solution(object):
    def permute(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        if not nums:      #sanity check
            return []
        
        res = []
        self.helper(nums, res, [])
        return res
    
    def helper(self, nums, res, item):   #explore
        if len(item) == len(nums):
            res.append(item[:]) #为什么要item[:]可以去研究研究python的参数传入
            return
        
        for num in nums:
            if num not in item:
                item.append(num)  #choose
                print item # put a print here just to give a hint of why we need to unchoose,you can see the item list growing and shrinking
                self.helper(nums, res, item)  #explore recursively
                item.pop()      #unchoose

再来看个例子：给定掷骰子的次数，再给定目标和，输出所有点数和等于目标和的组合

class Solution(object):
    def diceSum(self, num, sum):
        res = []
        self.helper(num, sum, res, [])
        return res
 
    def helper(self, num, sum, res, path):
        if num == 0:
            if sum == 0:
                res.append(path[:]) #path[:]就相当于Java里的new ArrayList(path),新内存
                return
            return
        for i in xrange(1,7):
            path.append(i)   #choose
            self.helper(num-1, sum-i, res, path)  #explore
            path.pop()    #unchoose

上面的程序可以解决问题，可是还不完美，我们可以进行剪枝优化，这里代码就不列出了。