java集合排序问题

Question

我使用简单的比较器并得到异常，但不知道该怎么办

这就是我的称呼:

try {
   Collections.sort(this.closePositions, new PositionComperator());
}
catch(Exception e) {
   e.printStackTrace();
}

这是比较器:

  public class PositionComperator implements Comparator<DataResponse> {

    @Override
    public int compare( DataResponse pos1, DataResponse pos2) {

        if (pos1.openTime >= pos2.openTime) {
            return 1;
        } 
        else {
            return -1;
        }// returning 0 would merge keys

    }

   }

这是异常(exception):

java.lang.IllegalArgumentException: Comparison method violates its general contract!
at java.util.TimSort.mergeLo(Unknown Source)
at java.util.TimSort.mergeAt(Unknown Source)
at java.util.TimSort.mergeCollapse(Unknown Source)
at java.util.TimSort.sort(Unknown Source)
at java.util.TimSort.sort(Unknown Source)
at java.util.Arrays.sort(Unknown Source)
at java.util.Collections.sort(Unknown Source)
at GTTask.RefreshIdentityHistory.call(RefreshIdentityHistory.java:59)
at GTTask.RefreshIdentityHistory.call(RefreshIdentityHistory.java:1)
at java.util.concurrent.FutureTask$Sync.innerRun(Unknown Source)
at java.util.concurrent.FutureTask.run(Unknown Source)
at java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source)
at java.lang.Thread.run(Unknown Source)

Answer 1

如果两个值x和y具有相同的openTime，则compare(x, y)和compare(y, x) 都会返回 1，这违反了 compare 的约定:

The implementor must ensure that sgn(compare(x, y)) == -sgn(compare(y, x)) for all x and y.

你还没有确保这一点。

您需要考虑当 openTime 值相同时您想要发生什么 - 要么返回 0，要么对哪个值应该出现在其他。例如，您是否可以进行一些二次比较？