LeetCode#46 Permutations

Problem Definition:

Given a collection of 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], and [3,2,1].

Solution:

1）借用 Next Permutation 算法（线性），从最小的开始，找下一个......

 1     # @param {integer[]} nums
 2     # @return {integer[][]}
 3     def permute(self, nums):
 4         nums.sort()
 5         pm=[nums[:]]
 6         cnt=reduce(lambda x, y: x*y, range(1, len(nums)+1))
 7         for _ in range(cnt-1):
 8             self.nextPermute(nums),
 9             pm+=nums[:],
10         return pm
11 
12     def nextPermute(self, nums):
13         n=len(nums)
14         if n<2:
15             return
16         i,j,pLeft,pRight=n-2,n-1,-1,n-1
17         while i>=0 and nums[i]>=nums[i+1]:
18             i-=1
19         if i>=0:
20             while j>i and nums[j]<=nums[i]:
21                 j-=1
22             #must be j>i
23             nums[i],nums[j]=nums[j],nums[i]
24         pLeft=i+1
25         while pLeft<pRight:
26             nums[pLeft],nums[pRight]=nums[pRight],nums[pLeft]
27             pLeft+=1
28             pRight-=1

 

2）上面的解法，实现nextPerte函数有很多的细节要考虑。这题用回溯法比较好，idea比较直接，实现起来容易，性能也好。

 1     # @param {integer[]} nums
 2     # @return {integer[][]}
 3     def permute(nums):
 4         pms=[]
 5         backTrack(nums, [], pms, len(nums))
 6         return pms
 7 
 8     def backTrack(nums, localArr, pms, n):
 9         if len(localArr)==n:
10             pms+=localArr[:],
11         else:
12             for i, e in enumerate(nums):
13                 newNums=nums[:i]+nums[i+1:]
14                 backTrack(nums[:i]+nums[i+1:], localArr+[e], pms, n)

 

posted @ 2015-08-04 18:56  曾可爱  阅读(115)  评论(0)  编辑  收藏  举报