首页 > 代码库 > 字符串的全排列
字符串的全排列
假定字符串S,以字符序列a1a2...an表示。例如: 对于字符串"abc", 全排列为cba bca bac cab acb abc
本文采用非递归方法分别给出Python版实现和C代码实现。
1 #!/usr/bin/python 2 import sys 3 4 def str2listuniq(s): 5 l = [] 6 for c in list(s): 7 if c not in l: 8 l.append(c) 9 return l 10 11 def insert_c2s(s, c): 12 l = [] 13 i = 0 14 while i <= len(s): 15 start = s[0:i] 16 end = s[i:] 17 e = start + c + end 18 l.append(e) 19 i += 1 20 return l 21 22 def list2allperms(l): 23 l_final = [l[0]] 24 for c in l[1:]: 25 l_cell = [] 26 for s in l_final: 27 l_cell.extend(insert_c2s(s, c)) 28 l_final = l_cell 29 return l_final 30 31 def main(argc, argv): 32 if argc != 2: 33 sys.stderr.write("Usage: %s <string>\n" % argv[0]) 34 sys.stderr.write(" e.g.: %s \"abc\"\n" % argv[0]) 35 return -1 36 37 if len(argv[1]) == 0: 38 sys.stderr.write("Oops, string is blank\n") 39 return -1 40 41 l_input = str2listuniq(argv[1]) 42 l_final = list2allperms(l_input) 43 if len(l_final) <= 24: 44 print ‘ ‘.join(l_final) 45 else: 46 print ‘ ‘.join(l_final[0:5]) + " ... " + l_final[-1] 47 print "\nThe total number of all permutation is %d" % len(l_final) 48 49 return 0 50 51 if __name__ == ‘__main__‘: 52 argc, argv = len(sys.argv), sys.argv 53 sys.exit(main(argc, argv))
测试结果:
$ ./foo.py a a The total number of all permutation is 1 $ ./foo.py ab ba ab The total number of all permutation is 2 $ ./foo.py abc cba bca bac cab acb abc The total number of all permutation is 6 $ ./foo.py abcd dcba cdba cbda cbad dbca bdca bcda bcad dbac bdac badc bacd dcab cdab cadb cabd dacb adcb acdb acbd dabc adbc abdc abcd The total number of all permutation is 24 $ ./foo.py abcde edcba decba dceba dcbea dcbae ... abcde The total number of all permutation is 120
用C代码实现比较复杂一些,因为要构造动态数组,wait for a while ...
字符串的全排列
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。