Question

我怀疑在列表中 4 teams之间实施抽奖我在下面有这个例子，并且它仅适用于结果不正确，代码有效，您可以点击链接http://ideone.com/bs9oAZ并点击此处

    String[] teams = new String[4];
    teams[0] = "BOSTON";
    teams[1] = "CLEVELAND";
    teams[2] = "SPURS";
    teams[3] = "WARRIORS";

    System.out.println("Number of teams:" + teams.length);

    System.out.print("\n[ ");
    for (String team : teams) {
        System.out.print(team + " ");
    }
    System.out.println("]");

    for (int i = 0; i < teams.length - 1; i++) {
        System.out.format("\n--- Match %02d---\n", i + 1);
        for (int j = 0; j < teams.length; j++) {
            for (int k = j + 1; k < teams.length - 1; k++) {
                System.out.println(teams[j] + " - " + teams[k]);
            }
        }
    }

结果是第一轮

Number of teams: 4
[ BOSTON  CLEVELAND SPURS WARRIORS ]

    --- Match 01---
BOSTON - CLEVELAND
BOSTON - SPURS
CLEVELAND - SPURS

--- Match 02---
BOSTON - CLEVELAND
BOSTON - SPURS
CLEVELAND - SPURS

--- Match 03---
BOSTON - CLEVELAND
BOSTON - SPURS
CLEVELAND - SPURS

以及 4 teams的正确结果，应该是第一轮

--- Match 01 ---
BOSTON - CLEVELAND
SPURS  - WARRIORS

--- Match 02 ---
BOSTON    - SPURS
CLEVELAND - WARRIORS

--- Match 03 ---
BOSTON    - WARRIORS
CLEVELAND - SPURS

如果我能解决第一轮，我将尝试第二轮。

任何建议？

Answer 1

你当然可以使用数组，但我确信使用ArrayLists会更容易。这是因为你正在将波士顿与＆＃34; X＆＃34;然后配对剩下的两个，其中ArrayList类有很多方便的方法。这里是您的代码（请注意，您仍然以字符串数组开头）：

    String[] teams = {"BOSTON", "CLEVELAND", "SPURS", "WARRIORS"};

    System.out.println("Number of teams:" + teams.length);

    System.out.print("\n[ ");
    for (String team : teams)
        System.out.print(team + " ");
    System.out.println("]");

    ArrayList<String> pair1, pair2;
    for (int i = 1; i < teams.length ; i++) 
    {
        pair1 = new ArrayList<String>(); // initializes this ArrayList
        for(int p = 0; p < teams.length ; p++) // adds the contents of "teams" 
            pair1.add(teams[p]);

        pair2 = new ArrayList<String>();
        pair2.add(pair1.remove(i)); // pair1 has 4 elements,
        pair2.add(pair1.remove(0)); // 2 of which are snatched and put in pair 2.
        pair2.add(0,pair2.remove(1)); //swaps the elements in pair2

        System.out.format("\n--- Match %02d---\n", i);
        System.out.println(pair2.get(0)+"-"+pair2.get(1));
        System.out.println(pair1.get(0)+"-"+pair1.get(1));
    }

正如您所看到的，这个过程非常复杂，甚至可能不是最好的过程。但到目前为止，这是我提出的最简单的问题。如果你想要一个使用数组，我可以编辑它并将其放入。

Answer 2

这就是我要做的事情。仅适用于偶数团队。如果您不理解它的任何部分，请随时问我。

public class Match {
      public static int factorial(int num) {
          int i = 1 ;
          while(num != 1) {
              i = i * num ;
              num-- ;
          }
          return i ;
      }
      public static int search(int[] array) {
          int var ;
          for(var = 0 ; var < array.length ; var++ ) {
             if(array[var] == 0) {
              return var ;
             }
          }
          return -1 ;
      }
      public static void main(String [] args) {
            String[] teams = new String[4];
            teams[0] = "BOSTON";
            teams[1] = "CLEVELAND";
            teams[2] = "SPURS";
            teams[3] = "WARRIORS";
            int [] opp = new int[4] ; // array that determines if a team has got an opposition team
            int comb = (factorial(teams.length)) / (2 * factorial(teams.length - 2)) ; //find no.of possible combinations
            int match = comb / (teams.length/2) ; // find number of matches
            for(int i = 0 ; i < opp.length ; i++) { //clear array
                opp[i] = 0 ;
            }
            System.out.println("Number of teams:" + teams.length);
            System.out.print("[ ");
            for (String team : teams) {
                System.out.print(team + " ");
            }
            System.out.print("]\n");
                for (int j = 1 ; j <= match ; j++) {
                   System.out.format("\n--- Match %02d---\n", j);
                   System.out.println(teams[0] + " - " + teams[j]); //set the first team and its opponent
                   opp[0] = 1 ; // first team has been set 
                   opp[j] = 1 ; // opponent has been set
                   for(int i = 2 ; i <= teams.length/2 ; i++) { //first pair has been set, start from second pair
                       int var1 = search(opp); // look for next team without opponent
                       opp[var1] = 1 ;
                       int var2 = search(opp); // look for next team without opponent
                       opp[var2] = 1 ;
                      System.out.println(teams[var1] + " - " + teams[var2]);
                   }
                   for(int i = 0 ; i < opp.length ; i++) { //clear the array
                        opp[i] = 0 ;
                    }
                }

      }
}

使用Java

2 个答案: