47. Permutations II 【M】【61】

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本文介绍了一种算法，用于找出包含重复数字的所有唯一排列。通过逐步添加数字并处理重复情况，确保结果中没有重复的排列。文章详细展示了算法的实现过程，并提供了Python代码示例。

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

For example,
[1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1].

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the basic idea is, to permute n numbers, we can add the nth number into the resulting List<List<Integer>> from the n-1 numbers, in every possible position.

For example, if the input num[] is {1,2,3}: First, add 1 into the initial List<List<Integer>> (let's call it "answer").

Then, 2 can be added in front or after 1. So we have to copy the List in answer (it's just {1}), add 2 in position 0 of {1}, then copy the original {1} again, and add 2 in position 1. Now we have an answer of {{2,1},{1,2}}. There are 2 lists in the current answer.

Then we have to add 3. first copy {2,1} and {1,2}, add 3 in position 0; then copy {2,1} and {1,2}, and add 3 into position 1, then do the same thing for position 3. Finally we have 2*3=6 lists in answer, which is what we want.

就是说，用 [1，1，3] 举例子

每次把一个数字放进去，每个位置都放一遍。那处理重复怎么办呢？遇见重复就break就行了。。

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

class Solution(object):
    def permuteUnique(self, nums):
        ans = [[]]
        for n in nums:
            new_ans = []
            for l in ans:
                #print ans
                for i in xrange(len(l)+1):
                    #print l[:i]+[n]+l[i:]
                    new_ans.append(l[:i]+[n]+l[i:])
                    if i<len(l) and l[i]==n: #i < len(l) 保证数组不越界
                        break
                    #handles duplication
            ans = new_ans
        return ans
            
            
        '''
        p = itertools.permutations(nums)
        res = set()
        for i in p:
            res.add(i)
        ress = []
        for i in res:
            ress += [list(i)]
        return ress
        
        '''