Leetcode篇：全排列

Posted 2021-01-20 恩zzq我是

---

@author: ZZQ
@software: PyCharm
@file: permute.py
@time: 2018/11/15 19:42
要求：给定一个没有重复数字的序列，返回其所有可能的全排列。
示例:
输入: [1,2,3]
输出:
[
[1,2,3],
[1,3,2],
[2,1,3],
[2,3,1],
[3,1,2],
[3,2,1]
]

import copy

"""
思路一： DFS，去掉不满足条件的排列
"""
class Solution():
    def __init__(self):
        pass

    def permute(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        ans = []
        temp_ans = []
        length = len(nums)
        cur_length = 0
        self.dfs(temp_ans, nums, cur_length, length, ans)
        return ans

    def dfs(self, temp_ans, nums, cur_length, length, ans):
        if cur_length == length:
            tt_ans = copy.deepcopy(temp_ans)
            ans.append(tt_ans)
        else:
            for i in range(length):
                if nums[i] not in temp_ans:
                    temp_ans.append(nums[i])
                    self.dfs(temp_ans, nums, cur_length+1, length, ans)
                    temp_ans.pop()

"""
思路二： 交换元素+DFS
"""
class Solution2():
    def __init__(self):
        pass

    def permute(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        nums_len = len(nums)
        ans = []
        if nums_len == 0 or nums == []:
            return []
        self.exchange(nums, 0, nums_len, ans) # 采用前后元素交换的办法，dfs解题
        return ans

    def exchange(self, nums, i, len, ans):
        if i == len-1:  # 将当前数组加到结果集中
            temp_list = []
            for j in range(len):
                temp_list.append(nums[j])
            ans.append(temp_list)
            return
        # 将当前位置的数跟后面的数交换，并搜索解
        for j in range(i, len):
            temp = nums[i]
            nums[i] = nums[j]
            nums[j] = temp
            self.exchange(nums, i+1, len, ans)
            temp = nums[i]
            nums[i] = nums[j]
            nums[j] = temp