46. Permutations (全排列)

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]
]

运用递归。 1234为例子

for  i in 1234:

　　1  +  234(的全排列)

　　2  +  134（的全排列）

　　3  +  124（的全排列）

　　4   + 123 （的全排列）

对应程序的17行

 class Solution(object):
     def __init__(self):
         self.res = []

     def permute(self, nums):
         """
         :type nums: List[int]
         :rtype: List[List[int]]
         """
         self.help(nums, 0, len(nums))

         return self.res

     def help(self, a, lo, hi):
         if(lo == hi):
             self.res.append(a[0:hi])
         for i in range(lo, hi):
             self.swap(a, i, lo)
             self.help(a, lo + 1, hi)
             self.swap(a, i, lo)
     def swap(self, a, i, j):
         temp = a[i]
         a[i] = a[j]
         a[j] = temp

非递归

 class Solution(object):

     def permute(self, nums):
         """
         :type nums: List[int]
         :rtype: List[List[int]]
         """
         nums = sorted(nums.copy())
         res = [nums.copy()]

         n = self.next_permute(nums)
         while True:

             if n is None:
                 break
             res.append(n)
             n = self.next_permute(n.copy())

         return res

     def next_permute(self, a):
         i = -1
         for index in range(0, len(a) - 1)[::-1]:
             if(a[index] < a[index + 1]):
                 i = index
                 break
         if i==-1:
             return None
         min_i = a.index(min([k for k in a[i+1:] if k >a[i]]))
         self.swap(a, i, min_i)
         an = a[:i + 1] + self.revers(a[i + 1:])
         return an

     def revers(self, x):
         for i in range(int(len(x) / 2)):
             temp = x[i]
             x[i] = x[len(x) - i - 1]
             x[len(x) - i - 1] = temp
         return x

     def swap(self, a, i, j):
         temp = a[i]
         a[i] = a[j]
         a[j] = temp