[英]Check if int[] contains an int
我的脚本陷入无限循环,我不知道为什么。 我有一个包含 40 个棋子的数组,需要将这些棋子随机放在棋盘上。 所以,我有一个随机数从数组中选择一个随机 pawn,但是如果 pawn 已经被选择,它必须选择一个新的随机数,但由于某种原因,最后一部分似乎有问题。 我不明白为什么。
Random rand = new Random();
int[] availablePawnsArray = {1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 8, 8, 9, 10, 11, 12, 12, 12, 12, 12, 12 };
// this array contains 40 integers
int[] chosenPawns = new int[40];
//this array contains the index numbers of already selected pawnsfrom the previous array
int counter = 0;
//counts how many pawns have been selected already
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 10; j++) {
//this refers to my board, 40 locations for my 40 pawns
int chosenPawn = rand.nextInt(40);
//a random numder from 0 to 40
boolean found = false;
//a boolean to say if i have already selected this pawn before or not
do {
for (int n : chosenPawns) {
if (n == chosenPawn) {
found = true;
chosenPawn = rand.nextInt(40);
} else {
found = false;
}
}
} while(found == true);
board[i][j].rank = availablePawnsArray[chosenPawn];
chosenPawns[counter] = chosenPawn;
counter++;
}
}
您可以有两个数组,第二个保留选定的整数,然后在第二个数组中循环检查是否有任何数字等于给定一个返回 false 或 true。
int [] selectedInts = new int[40];
boolean contains(int num) {
for (int i = 0 ; i < selectedInts.length; i++) {
if (i == num) return true;
}
return false;
}
你也可以使用像
Arrays.asList().contains(yourInt);
您可以通过修改随机数选择来简化这一过程,就像洗一副纸牌一样。 这是Fisher-Yates 的一个轻微变体。 我相信这应该始终在线性时间内起作用。
Random rand = new Random();
int[] availablePawnsArray = {
1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6,
6, 6, 6, 7, 7, 7, 8, 8, 9, 10, 11, 12, 12, 12, 12, 12, 12
};
int start = 39;
for (int i = 0; i < 4; i++) {
for (int k = 0; k < 10; k++) {
int chosenPawn = rand.nextInt(start + 1);
// chose the pawn
board[i][k].rank = availablePawnsArray[chosenPawn];
// copy the pawn from the end of the list to the
// chosen pawn location.
availablePawnsArray[chosenPawn] = availablePawnsArray[start];
// update the random number to ignore the last slot
// in the array (the pawn in that slot has
// been moved to occupy the chosenPawn's location)
start--;
}
}
for (int i = 0; i < 4; i++) {
for (int k = 0; k < 10; k++) {
System.out.print(board[i][k].rank + " ");
}
System.out.println();
}
}
除非您需要将其作为练习来实现,否则您可以使用内置的shuffle
方法,将可用 pawn 的数组包装在一个列表中:
Collections.shuffle(Arrays.asList(availablePawnsArray));
for (int i = 0, k = 0; i < 4; i++)
for (int j = 0; j < 10; j++, k++)
board[i][j].rank = availablePawnsArray[k];
selectedPawns 数组应该只搜索到计数器。 当几乎全部都被使用时,使用重复随机来寻找一个空闲位置,可能需要数千步才能找到最后的最后一个空闲位置。 甚至不能保证找到空位。
简单地将随机位置作为起点来找到第一个空闲的棋子,从...... 36, 37, 38, 39, 0, 1, ...开始循环。
int[] pawnsArray = {1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4,
5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 8, 8, 9, 10, 11, 12, 12, 12, 12, 12, 12 };
// this array contains 40 integers
boolean[] pawnsChosen = new boolean[40];
// this array tells wether the i'th pawn was chosen.
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 10; j++) {
int pawnIndex = rand.nextInt(40);
while (pawnsChosen[pawnIndex]) {
pawnIndex = (pawnIndex + 1) % 40;
}
pawnsChosen[pawnIndex] = true;
board[i][j].rank = availablePawnsArray[pawnIndex];
}
}
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