package com.kpp; /** * 求字符串的全排列 * 递归的思想 * 比如 abcde 先求出abcd的全排列,然后将e分别插入全排列的5个位置 * a 全排列 a * ab 全排列 ab ba * abd 全排列即是 cab acb abc cba bca bac * * @author kpp * */ public class QuanPaiLie { /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub String str = "abcde"; //System.out.println(str.substring(0, 2)); String[] rs = getQuanPaiLie(str); if(rs != null){ int count = 0; for(int i = 0;i < rs.length;i++){ if(rs[i]!=null){ System.out.println(rs[i]); count++; } } System.out.println("count: "+count); } } /** * 求一个字符串的全排列 * @param str * @return */ public static String[] getQuanPaiLie(String str){ String[] rs = new String[1000]; if(str == null||str.isEmpty()){ return null; } int strLen = str.length(); if(strLen == 1){ rs[0] = str; }else if(strLen == 2){ rs[0] = str; rs[1] = ""+str.charAt(1)+str.charAt(0); }else if(strLen > 2){ char c = str.charAt(strLen-1); String strBefore = str.substring(0,strLen-1); String[] tmpRsArr = getQuanPaiLie(strBefore); int count = 0; for(int i = 0;i < tmpRsArr.length;i++){ String tmpRs = tmpRsArr[i]; if(tmpRs != null){ for(int k = 0;k < tmpRs.length()+1;k++){ rs[count++] = tmpRs.substring(0,k)+c+tmpRs.substring(k,tmpRs.length()); } } } } return rs; } }
时间: 2024-11-10 01:10:23