Backtracking 编程范式python

 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。