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当我运行它时,合并排序没有发生,我无法弄清楚问题出在哪里。存在 int brelem 和 SortThread 任务参数只是因为我需要它们正确地覆盖父类的方法,因此它们在这种情况下不起任何作用。

class MergeSort extends Sort {

@Override
ArrayList sort(ArrayList<Integer> a, int brelem, SortThread task) {
    if (a.size() > 1) {

        ArrayList<Integer> firstHalf = new ArrayList<>(a.subList(0, a.size() / 2));

        firstHalf.subList(0, a.size() / 2);
        firstHalf=sort(firstHalf, a.size() / 2, task);

        ArrayList<Integer> secondHalf = new ArrayList<>(a.subList(a.size()/2, a.size()));

        secondHalf=sort(secondHalf, secondHalf.size(), task);

        ArrayList<Integer> temp = merge(firstHalf, secondHalf);
        a.subList(0, temp.size());
    }
    return a;
}

private static ArrayList merge(ArrayList<Integer> list1, ArrayList<Integer> list2) {
    ArrayList<Integer> temp = new ArrayList<>();

    int current1 = 0;
    int current2 = 0;
    int current3 = 0;

    while (current1 < list1.size() && current2 < list2.size()) {
        if (list1.get(current1) < list2.get(current2)) {
            temp.set((current3++), list1.get(current1++));
        } else {
            temp.set((current3++), list2.get(current2++));
        }
    }

    while (current1 < list1.size()) {
        temp.set((current3++), list1.get(current1++));
    }

    while (current2 < list2.size()) {
        temp.set((current3++), list2.get(current2++));
    }

    return temp;
}

}

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1 回答 1

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除其他外,您正在返回a,而不是temp。据我所知,您永远不会修改a.

也许你对做什么感到困惑List.subList()。它返回列表的视图,因此无论您在sort倒数第二条语句中使用它做什么都将无效,因为您忽略了返回值。

于 2012-12-06T20:26:27.777 回答