将元素插入二维数组中的下一个空索引的最有效方法是什么？

Question

我正在寻找一种不太复杂的方法（如果可能）将元素插入到二维数组的下一个空垂直索引，同时随机循环遍历水平索引。

例如：

制作一张表格，显示来自 8 支球队的 32 名球员，随机分配到 4 个回合中。所以，我们希望每轮有 8 名玩家。

考虑以下内容：roundsTable[x][y] 是表格，而 x 代表每个玩家参加的回合，y 代表每个回合中的玩家。玩家将从另一个二维数组players[t][p] 中收集，并将按索引顺序 t(0,1,2,3,4,5,6,7) p(0,1,2,3) 收集。但是，它们将在每次随机循环x 中存储在roundsTable[][] 中。问题是，我怎样才能在每个被选中的随机x 中将每个玩家放在y 的下一个空白位置。

//playersRoundOrder[] is 32 long and contains the random rounds in a range of 1-4. It looks something like that: {1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,3,3,3...} then I shuffle it to become something like this: {4,1,1,3,2,3,4...}. Then, I take the first index of that array and I set it as the round for the first index of player in the first team, then the second index to the second player of the first team...etc 
private void example(int[] playersRoundOrder) { 
    String[][] roundsTable = new String[4][8]; //The table that I will use to store each player based on their round. So, the player whose round is 3, will be stored at roundsTable[3][*Here is where I need the solution. I want to insert each player in round three in the next available spot etc.*]
    int t = 0; //count the index of each team. So, it will only increase when I set a random round for each player in the first team.
    int p = 0; //count the index of each player in t. So, it will increase after each 1 loop, and it returns to zero when t increase.

    for (int i = 0; i < playersRoundOrder.length; i++) {// loop 32 times
        //String players[][] is the other two dimensional array that contains 32 players
        //(8 teams - 4 players for each)
        if ((p + 1) == players[t].length) { //If there is no more players in this team t, go to the next team and start from the index of the first player
            roundsTable[playersRoundOrder[i]] [**PROBLEM**] = players[++t][p = 0];
        } else { //If there is still players on the team, go to the next player
          roundsTable[playersRoundOrder[i]] [**PROBLEM**] = players[t][p++];
        }
    }
}

Answer 1

如果我正确理解你的问题，你有以下持有球员的名字：

String players[8][4];  // first dim team, second dim player number

你想在下面填写每轮所有玩家的名字：

String roundsTable[4][8]; // first dim round, second dim participant in round

这样

1. 每个玩家出现一次
2. 每队每轮有一名球员
3. 玩家在每一轮中出现的顺序是随机的
4. 分配到该轮每个位置的团队是随机的

如果我错了，请纠正我。如果没有，那么这是一个潜在的解决方案：

String[][] generateRoundsTable(String[][] players) {
    String[][] roundsTable = new String[ROUND_COUNT][TEAM_COUNT];
    List<Integer>[] assignment = new List<>[ROUND_COUNT];
    for (int r = 0; r < ROUND_COUNT; r++) {
        assignment[r] = IntStream.range(0, TEAM_COUNT).collect(Collectors.toList());
       Collections.shuffle(assignment[r]);
    }
    for (int t = 0; t < TEAM_COUNT; t++) {
        List<String> team = Arrays.asList(players[t]);
        Collections.shuffle(team);
        for (int r = 0; r < ROUND_COUNT; r++) {
            roundsTable[r][assignment[r].get(t)] = team.get(r);
        }
    }
    return roundsTable;
}