冒泡按特定列值对2D数组行进行排序

时间:2014-04-09 21:04:35

标签: java arrays sorting

我有一个二维数组,我想通过数组第二列值对它进行冒泡排序。

我从用户那里获取Arrival timeService time值,并希望按数组第二列值(Arrival time)对其进行冒泡排序。

第一列是流程编号。

   static int[][] atst = new int[5][5];
    for (int i = 0; i < atst.length; i++) {
        System.out.print("Arrival time for process " + i + ": ");
        atst[i][1] = in.nextInt();
    }

    for (int i = 0; i < atst.length; i++) {
        System.out.print("Enter service Times for process " + i + ": ");
        atst[i][2] = in.nextInt();
    }

   System.out.println("Before sorting: " + Arrays.deepToString(atst));

    for (int i = 0; i < atst.length; i++) {
        for (int j = 1; j < (atst.length - 1); j++) {
            if (atst[j - 1][1] > atst[j][1]) {     // Then swap!
                int[] tempRow = atst[j - 1];
                atst[j - 1] = atst[j];
                atst[j] = tempRow;
            }
        }
    }

    System.out.println("After sorting :" + Arrays.deepToString(atst));

public static void swapRows(int[][] array, int rowA, int rowB) {
    int[] tempRow = array[rowA];
    array[rowA] = array[rowB];
    array[rowB] = tempRow;
}

swapRows方法有效,但它不完全排序数组。

结果:

Arrival time for process 0: 5
Arrival time for process 1: 4
Arrival time for process 2: 3
Arrival time for process 3: 2
Arrival time for process 4: 1

Enter service Times for process 0: 2
Enter service Times for process 1: 3
Enter service Times for process 2: 4
Enter service Times for process 3: 5
Enter service Times for process 4: 2

Before sorting: [[0, 5, 2, 0, 0], [1, 4, 3, 0, 0], [2, 3, 4, 0, 0], [3, 2, 5, 0, 0], [4, 1, 2, 0, 0]]
After sorting :[[3, 2, 5, 0, 0], [2, 3, 4, 0, 0], [1, 4, 3, 0, 0], [0, 5, 2, 0, 0], [4, 1, 2, 0, 0]]

结果应该是这样的:

[[4, 1, 2, 0, 0],[3, 2, 5, 0, 0],[2, 3, 4, 0, 0],[1, 4, 3, 0, 0],[0, 5, 2, 0, 0]]

1 个答案:

答案 0 :(得分:1)

在更新的代码中,内部循环的边界不正确:

for (int j = 1; j < (atst.length - 1); j++) {

你通过在这里减去1来排除你的最后一个元素,这就是除了最后一个元素之外数组的其余部分被排序的原因。应该是:

for (int j = 1; j < atst.length; j++) {