discard ( i ) # remove(i) will throw exception if i not existingĭfs ( nums, , res, visited ) return res #Ĭlass Solution : def permute ( self, nums : List ) -> List ]: def dfs ( i ): if i = n : ans. append ( nums ) dfs ( nums, path, res, visited ) path. Note: The reason that we have only 2 choices instead of 3, is that there is a duplicate in the given input. Suppose that we pick the number 1, now the remaining numbers would become 1, 2. append ( path + ) return for i in range ( 0, len ( nums )): if i not in visited : visited. Given the input of 1, 1, 2, at the first stage, we have 2 choices to pick a number as the first number in the final permutation, i.e. Cannot retrieve contributors at this time. ''' class Solution : def permute ( self, nums : List ) -> List ]: res = visited = set () def dfs ( nums, path, res, visited ): if len ( path ) = len ( nums ): res. Permutations.md Go to file Go to file T Go to line L Copy path Copy This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. * remove() removes the specified element from the set if it is present, but raises a KeyError exception if the element is not present. * discard() removes the specified element from the set if it is present, but does nothing if the element is not present. In Python 3, both the discard() and remove() methods of a set object are used to remove an element from the set, but there is one key difference: pop ( index ) ans = tmp_list return ans ''' append ( new_perm ) res = new_res return res class Solution : # iterative, single_perm.insert(index, num)ĭef permute ( self, nums : List ) -> List ]: ans = ] if nums is None or len ( nums ) = 0 : return ans for num in nums : tmp_list = for single_perm in ans : for index in range ( len ( single_perm ) + 1 ): single_perm. > del my_list # remove the element at index 2ĭef permute ( self, nums : List ) -> List ]: res = ] if nums is None or len ( nums ) = 0 : return ans for num in nums : new_res = for perm in res : for i in range ( len ( perm ) + 1 ): new_perm = perm + + perm new_res. Remove an element by its value from a setĬannot remove an element by index from a set in Python3 Import import import public class Permutations Then according to the above conclusion, it is actually obtained by adding a3 at different positions on the basis of a1a2 and a2a1. At this time, there are six kinds of full arrays, namely a1a2a3, a1a3a2, a2a1a3, a2a3a1, a3a1a2, and a3a2a1. Return a list of all possible strings we could create. Then, considering the relationship with the above situation at this time, you can find that it is actually added in the two positions before and after a1. Letter Case Permutation - Given a string s, you can transform every letter individually to be lowercase or uppercase to create another string. When n=2, there is a1a2 in the array at this time, and there are two kinds of full arrays, a1a2 and a2a1. When n=1, there is only one number a1 in the array, and there is only one total permutation, which is a1 See the image above for clarification.ĭo this for all the cases and it will generate all possible permutations of the given array.Given an array nums of distinct integers, return all the possible permutations. Fixing the second position automatically fixes the third position. In the first column of second-level 1 is fixed at the first position, in the second column 2 is fixed at the first position and in the third column 3 is fixed at the first position.Īfter fixing an element at the first position, fix an element at the second position, consider the case in the second level and the first column, that is,, 1 is fixed at the first position, so we have 2 choices for the second position that is either 2 or 3. The image below the second level represents this situation. Explanation for Leetcode problem Permutationsįix an element in the first position, we have three choices 1, or 2, or 3.
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