LeetCode 46. Permutations

Posted 2020-09-21 叶常绿

原题

Given a collection of distinct numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:

[
  [1,2,3],
  [1,3,2],
  [2,1,3],
  [2,3,1],
  [3,1,2],
  [3,2,1]
]

解题思路

递归：递归的方法，创建一个visit判断此值是否已经添加过，每一层不断地循环，加入没有被访问的元素，直到最后结果的长度满足要求加入答案中

class Solution(object):
    def permute(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        if nums == None:
            return []
        visit = [0 for i in range(len(nums))]
        self.ret = []
        self._permute(nums, visit, 0, [])
        return self.ret
        
    def _permute(self, nums, visit, count, ret):
        if count == len(nums):
            self.ret.append(ret)
            return
        for i in xrange(len(nums)):
            if visit[i] == 0:
                # ret += [nums[i]]  容易出错，如果加入这句后面需要还原，不然影响后面的循环
                visit[i] = 1
                self._permute(nums, visit, count+1, ret+[nums[i]])
                visit[i] = 0

非递归：跟之前求subsets的思路类似（不过这个是求重复项，sbusets是求非重复），也是从一个个元素一层层加上去，只不过加入结果的时候必须满足长度要求

class Solution(object):
    def permute(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        if nums is None:
            return []
        result = []
        self.helper(nums, [], result)
        return result

    def helper(self, List, path, result):
        if len(path) == len(List):
            result.append(path)
            return
        
        # 跟subsets思路差不多，不过这个是重复，sbusets是非重复
        # 也是从一个个元素一层层加上去，只不过加入结果的时候必须满足长度要求
        for i in range(len(List)):
            # 不重复加入同一元素
            if List[i] in path:
                continue
            # path.append(List[i])
            self.helper(List, path + [List[i]], result)
            # path.pop()