Permutations全排列演算法研究
阿新 • • 發佈:2021-10-10
來至於Codewars題目Permutations
在這個kata中,您必須建立輸入字串的所有排列,並刪除重複項(如果存在)。這意味著,您必須以所有可能的順序從輸入中洗牌所有字母。
示例:
permutations('a');#['a']
permutations('ab');#['ab','ba']
permutations('aabb');#['aabb','abab','abba','baab','baba','bbaa']
排列的順序無關緊要。
我的解法
def permutations(nums): res = list(set(permute(nums))) return res def permute(nums): res = [] if len(nums) > 1: all_list = list(nums) for i in range(len(all_list)): heat = all_list.pop(i) res.extend([heat + one for one in permute(''.join(all_list))]) all_list.insert(i, heat) else: res.extend(nums) return res
網友的解法
import itertools
def permutations(string):
return list("".join(p) for p in set(itertools.permutations(string)))
itertools.permutations工具原始碼
def permutations(iterable, r=None): # permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC # permutations(range(3)) --> 012 021 102 120 201 210 pool = tuple(iterable) n = len(pool) r = n if r is None else r if r > n: return indices = list(range(n)) cycles = list(range(n, n-r, -1)) yield tuple(pool[i] for i in indices[:r]) while n: for i in reversed(range(r)): cycles[i] -= 1 if cycles[i] == 0: indices[i:] = indices[i+1:] + indices[i:i+1] cycles[i] = n - i else: j = cycles[i] indices[i], indices[-j] = indices[-j], indices[i] yield tuple(pool[i] for i in indices[:r]) break else: return
附贈itertools.combinations工具原始碼
def combinations(iterable, r): # combinations('ABCD', 2) --> AB AC AD BC BD CD # combinations(range(4), 3) --> 012 013 023 123 pool = tuple(iterable) n = len(pool) if r > n: return indices = list(range(r)) yield tuple(pool[i] for i in indices) while True: for i in reversed(range(r)): if indices[i] != i + n - r: break else: return indices[i] += 1 for j in range(i+1, r): indices[j] = indices[j-1] + 1 yield tuple(pool[i] for i in indices)
permutations和combinations都是得到一個迭代器。
combinations方法重點在組合,permutations方法重在排列。