比较法违反其一般契约-- TimSort.mergeLo错误

Question

我在这里看到了与Timsort错误相关的各种问题，并在此基础上应用了代码更改。但它们都没有给我带来结果。请帮助我解决我的错误。代码：

    public int compare(Record record1, Record record2) {
    int indexOfRecord1=records.indexOf(record1);
    int indexOfRecord2=records.indexOf(record2);

    Integer sortIndexOfRecord1=indexToSortIndexMap.get(indexOfRecord1);
    Integer sortIndexOfRecord2=indexToSortIndexMap.get(indexOfRecord2);

    //Option 1: Throws same Timsort error:
    if(sortIndexOfRecord1!=null && sortIndexOfRecord2!=null){
        return sortIndexOfRecord1.compareTo(sortIndexOfRecord2);
    }else{
        return (sortIndexOfRecord1==null && sortIndexOfRecord2==null)?0:(sortIndexOfRecord1==null)?-1:1;
    }
   //Reframed the above logic like below. I got the same error
   //Option 2: Throws same Timsort error:
   /*
    if(sortIndexOfRecord1!=null && sortIndexOfRecord2!=null){
        if(sortIndexOfRecord1.intValue() == sortIndexOfRecord2.intValue())
            return 0;
        else if(sortIndexOfRecord1.intValue() < sortIndexOfRecord2.intValue())
            return -1;
        else if(sortIndexOfRecord1.intValue() > sortIndexOfRecord2.intValue())
            return 1;
    }else if(sortIndexOfRecord1==null && sortIndexOfRecord2==null){
        return 0;
    }else if(sortIndexOfRecord1==null && sortIndexOfRecord2!=null){
        return -1;
    }else if(sortIndexOfRecord1!=null && sortIndexOfRecord2==null){
        return 1;
    }
    return 0;
    */


    //Option 3: Throws same Timsort error:
    /*
    int index1 = -1;
    int index2 = -1;
    if(record1 != null && record2 != null){
        index1 = indexToSortIndexMap.get(indexOfRecord1)!=null?indexToSortIndexMap.get(indexOfRecord1):-1 ;
        index2 = indexToSortIndexMap.get(indexOfRecord2)!=null?indexToSortIndexMap.get(indexOfRecord2):-1 ;
    }else if(record1 != null && record2 == null){
        index1 = indexToSortIndexMap.get(indexOfRecord1)!=null?indexToSortIndexMap.get(indexOfRecord1):-1 ;
    }else if(record1 == null && record2 != null){
        index2 = indexToSortIndexMap.get(indexOfRecord2)!=null?indexToSortIndexMap.get(indexOfRecord2):-1 ;
    }else{
        index1 = -1;
        index2 = -1;
    }

    if (index1 > index2){
        return +1;
    }else if (index1 < index2){
        return -1;
    }else{
        return 0;
    }
    */

以上所有情况都会产生以下错误：

    Caused by: 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)

Answer 1

这三个选项似乎都很好。但是，您可以尝试将它们替换为标准比较器，以确保：

return Comparator.nullsFirst(Integer::compare)
                 .compare(sortIndexOfRecord1, sortIndexOfRecord2);

一种可能的答案是，您实际上是在对用于获取记录索引的records列表进行排序。这真是个坏主意。