#coding:utf-8 def permutation(inStr, pos, parentData): if len(inStr) == 0: return if len(inStr) == 1: print "{" + inStr + "}" return # here we need a new buffer to avoid to pollute the other nodes. buffer = [] buffer.extend(parentData) # choose the element buffer.append(inStr[pos]) # get the remnant elements. subStr = kickChar(inStr, pos) # got one of the result if len(subStr) == 1: buffer.extend(subStr) print buffer return # here we use loop to choose other children. for i in range(len(subStr)): permutation(subStr, i, buffer) # a simple method to delete the element we choose def kickChar(src, pos): srcBuf = [] srcBuf.extend(src) srcBuf.pop(pos) return srcBuf def main(): input = [‘A‘,‘B‘,‘C‘,‘D‘] for i in range(len(input)): permutation(input, i, []) main()
输出：
C:\Python27\python.exe D:/work/search/3.py [‘A‘, ‘B‘, ‘C‘, ‘D‘] [‘A‘, ‘B‘, ‘D‘, ‘C‘] [‘A‘, ‘C‘, ‘B‘, ‘D‘] [‘A‘, ‘C‘, ‘D‘, ‘B‘] [‘A‘, ‘D‘, ‘B‘, ‘C‘] [‘A‘, ‘D‘, ‘C‘, ‘B‘] [‘B‘, ‘A‘, ‘C‘, ‘D‘] [‘B‘, ‘A‘, ‘D‘, ‘C‘] [‘B‘, ‘C‘, ‘A‘, ‘D‘] [‘B‘, ‘C‘, ‘D‘, ‘A‘] [‘B‘, ‘D‘, ‘A‘, ‘C‘] [‘B‘, ‘D‘, ‘C‘, ‘A‘] [‘C‘, ‘A‘, ‘B‘, ‘D‘] [‘C‘, ‘A‘, ‘D‘, ‘B‘] [‘C‘, ‘B‘, ‘A‘, ‘D‘] [‘C‘, ‘B‘, ‘D‘, ‘A‘] [‘C‘, ‘D‘, ‘A‘, ‘B‘] [‘C‘, ‘D‘, ‘B‘, ‘A‘] [‘D‘, ‘A‘, ‘B‘, ‘C‘] [‘D‘, ‘A‘, ‘C‘, ‘B‘] [‘D‘, ‘B‘, ‘A‘, ‘C‘] [‘D‘, ‘B‘, ‘C‘, ‘A‘] [‘D‘, ‘C‘, ‘A‘, ‘B‘] [‘D‘, ‘C‘, ‘B‘, ‘A‘] Process finished with exit code 0
代码借鉴于http://airu.iteye.com/blog/1930391的java代码
原文：http://www.cnblogs.com/coolor/p/6747218.html