两种方法-用1、2、2、3、4、5这六个数字,用java写一个main函数,打印出所有不同的排列,如:512234、412325等.要求:"4"不能在第三位 - 代码物语-Let code talk - ITeye技术网站


两种方法-用1、2、2、3、4、5这六个数字,用java写一个main函数,打印出所有不同的排列,如:512234、412325等.要求:"4"不能在第三位 - 代码物语-Let code talk - ITeye技术网站
  1.      * 题目:用1、2、2、3、4、5这六个数字,用java写一个main函数,打印出所有不同的排列,如:512234、412325等. 
  2.      * 要求:"4"不能在第三位,"3"与"5"不能相连 
  3.      * A(6,6)-A(5,5)-2*5*A(4,4)+2*3*A(3,3)=396,396/2=198 
  4.      * two solutions: 
  5.      * 1.Permutation 
  6.      * 2.Graph,depthFirst 
  1. public class Graph {  
  2.     private static final int[] DATA={1,2,2,3,4,5};  
  3.     private static final int LENGTH=DATA.length;  
  4.     private boolean[] visited;  
  5.     private int[][] matrix;  
  6.     private StringBuilder resultString;  
  7.     private Set<String> resultSet;//use 'Set' to reject duplicate string.  
  8.       
  9.     public static void main(String[] args) {  
  10.         Graph graph=new Graph();  
  11.         graph.initial();  
  12.         for(int i=0;i<LENGTH;i++){  
  13.             graph.depthFirst(i);//start from 1,2,2,3,4,5,find their corresponding DFS  
  14.         }  
  15.         graph.print();  
  16.     }  
  17.   
  18.     public void initial(){  
  19.         resultString=new StringBuilder();  
  20.         resultSet=new HashSet<String>();  
  21.         int n=LENGTH;  
  22.         visited=new boolean[n];  
  23.         matrix=new int[n][n];  
  24.         for(int i=0;i<n;i++){  
  25.             for(int j=0;j<n;j++){  
  26.                 if(i==j){  
  27.                     matrix[i][j]=0;  
  28.                 }else{  
  29.                     matrix[i][j]=1;  
  30.                 }  
  31.             }  
  32.         }  
  33.         //"3"与"5"不能相连  
  34.         matrix[3][5]=0;  
  35.         matrix[5][3]=0;  
  36.     }  
  37.       
  38.     public void depthFirst(int origin){  
  39.         //case 1.resultString includes DATA[origin]  
  40.         resultString.append(DATA[origin]);  
  41.         visited[origin]=true;  
  42.         if(resultString.length()==LENGTH){  
  43.             boolean ok=resultString.charAt(2)!='4';//"4"不能在第三位  
  44.             if(ok){  
  45.                 resultSet.add(resultString.toString());  
  46.             }  
  47.         }  
  48.         for(int i=0;i<LENGTH;i++){  
  49.             if(!visited[i]&&matrix[origin][i]==1){  
  50.                 depthFirst(i);  
  51.             }else{  
  52.                 continue;  
  53.             }  
  54.         }  
  55.         //case 2.resultString don't include DATA[origin]  
  56.         resultString.deleteCharAt(resultString.length()-1);//remove DATA[origin]  
  57.         visited[origin]=false;  
  58.     }  
  59. }  


  1. public class Perm {  
  2.     public static final int BAD_INDEX = 3;  
  3.     public static final int BAD_VALUE = 4;  
  4.     public static final int FIRST_VALUE = 3;  
  5.     public static final int SECOND_VALUE = 5;  
  6.     /*use 'Set' to reject duplicate string.Maybe we should do this at the very beginning(create the string),but how? 
  7. I google it,and I find this: 
  8. 1.let data = { 1, 2, 6, 3, 4, 5 }; 
  9. 2.get all the permutation of 'data',but only store the strings which match "str.matches("^.*6.*2.*$")" (or str.matches("^.*2.*6.*$")) 
  10. 3.str.replace('6','2') 
  11.         */  
  12.     private Set<String> resultSet=new HashSet<String>();  
  13.       
  14.     public static void main(String[] args) {  
  15.         Perm p = new Perm();  
  16.         int[] data = { 122345 };  
  17.         p.perm(data, 0, data.length - 1);  
  18.         Set<String> set=p.getResultSet();  
  19.         for(String str :set){  
  20.             System.out.println(str);  
  21.         }  
  22.         System.out.println(set.size());  
  23.           
  24.     }  
  25.   
  26.     //find all possible combination  
  27.     public void perm(int[] data, int begin, int end) {  
  28.         if (data == null || data.length == 0) {  
  29.             return;  
  30.         }  
  31.         if (begin == end) {  
  32.             boolean ok = check(data);//exclude the 'bad' string  
  33.             if (ok) {  
  34.                 String str=stringOf(data);  
  35.                 resultSet.add(str);  
  36.             }  
  37.         }  
  38.         for (int i = begin; i <= end; i++) {  
  39.             swap(data, begin, i);  
  40.             perm(data, begin + 1, end);  
  41.             swap(data, begin, i);  
  42.         }  
  43.     }  
  44.   
  45.     //exclude the 'bad' string--"4"不能在第三位,"3"与"5"不能相连  
  46.        //we can also use regular expression:(!str.matches("^..4.*$")&&!str.matches("^.*((35)|(53)).*$")&&str.matches("^.*2.*6.*$"))  
  47.     public boolean check(int[] data) {  
  48.         if (data == null || data.length == 0) {  
  49.             return false;  
  50.         }  
  51.         for (int i = 0, len = data.length; i < len - 1; i++) {  
  52.             if (data[i] == FIRST_VALUE && data[i + 1] == SECOND_VALUE  
  53.                     || data[i + 1] == FIRST_VALUE && data[i] == SECOND_VALUE) {  
  54.                 return false;  
  55.             }  
  56.             if (i + 1 == BAD_INDEX && data[i] == BAD_VALUE) {  
  57.                 return false;  
  58.             }  
  59.         }  
  60.         return true;  
  61.     }  
  62.   
  63.     //int[] data = { 1, 2, 2, 3, 4, 5 }-->"122345"  
  64.     public String stringOf(int[] x){  
  65.         StringBuilder sb=new StringBuilder();  
  66.         for(int i=0,len=x.length;i<len;i++){  
  67.             sb.append(x[i]);  
  68.         }  
  69.         return sb.toString();  
  70.     }  
  71.       
  72.     public void swap(int[] x, int i, int j) {  
  73.         int tmp = x[i];  
  74.         x[i] = x[j];  
  75.         x[j] = tmp;  
  76.     }  
  77.       
  78.     public Set<String> getResultSet(){  
  79.         return resultSet;  
  80.     }  
  81. }  
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