47. Permutation II Leetcode Python
阿新 • • 發佈:2019-01-25
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]
.
generally finding all permutation will take N! time based on T(n) = n*T(n-1)
this approach we can trim some of the cases, example: num[i] == num[i-1] but we have not yet visit num[i - 1]
so it would be a little less than n! but still np
def backtrack(num, res, val, visit): if len(val) == len(num) and : res.append(val) for i in range(len(num)): if i > 0 and visit[i - 1] == False and num[i - 1] == num[i]: continue if visit[i] == False: visit[i] = True backtrack(num, res, val+[num[i]], visit) visit[i] = False def permuteUnique(num): visit = [False for i in range(len(num))] res = [] val = [] num.sort() backtrack(num, res, val, visit) return res num = [1, 3, 1] print permuteUnique(num)