英文:
java.lang.IllegalArgumentException: Comparison method violates its general contract, only in particular test case
问题
这是我的比较函数：
Arrays.sort(intervals, (a, b) -> {
return (a[0] > b[0]) ? 1 : ((a[0] < b[0]) ? -1 : ((a[1] > b[1]) ? 1 : -1));
});
我认为我已经明确地为每个条件指定了返回值，当a和b是大小为2的数组时，但我不知道为什么在一个测试用例中出现错误，而其他测试用例则正常运行：
java.lang.IllegalArgumentException: Comparison method violates its general contract:
at line 903, java.base/java.utili.TimSort.mergeHi
at line 520, java.base/java.utili.TimSort.mergeAt
at line 448, java.base/java.utili.TimSort.mergeCollapse
at line 245, java.base/java.utili.TimSort.sort
at line 1442, java.base/java.utili.Arrays.sort
at line 3, Solution.merge
at line 54, __DriverSolution__.__helper__
at line 84, __Driver__.main
英文:
Here is my comparator function:
Arrays.sort(intervals, (a, b)->{
return (a[0] > b[0]) ? 1 : ((a[0] < b[0]) ? -1 : ((a[1] > b[1]) ? 1 : -1));
});
I think I've clearly specified return value for each condition in a case when a and b are arrays of size 2, but I don't know why I'm getting this error in a single test case while others run properly:
java.lang.IllegalArgumentException: Comparison method violates its general contract:
at line 903, java.base/java.utili.TimSort.mergeHi
at line 520, java.base/java.utili.TimSort.mergeAt
at line 448, java.base/java.utili.TimSort.mergeCollapse
at line 245, java.base/java.utili.TimSort.sort
at line 1442, java.base/java.utili.Arrays.sort
at line 3, Solution.merge
at line 54, __DriverSolution__.__helper__
at line 84, __Driver__.main
答案1
得分: 1
如果你未能涵盖一个情况，你的代码甚至无法编译。
就目前而言，你的代码可以编译，但它并不满足比较方法的要求。这是一个不同的问题。
以下是一般约定：
- 如果a < b且b < c，则a < c。
- 如果a == b，则compare(a, b) 必须返回0。
- 如果a < b，则b > a。
- 如果你愿意，可以抛出异常。
你已经违反了这个约定：如果我将相同的值传递给你的比较方法，最后一个情况会发生，并且返回-1，这意味着：“a在a下面”，这违反了规则。
将最后一个节点（-1）展开为：(a[1] < b[1]) ? -1 : 0。
英文:
If you had failed to cover a case, your code wouldn't even compile.
As is, your code compiles, but it doesn't fulfill the requirements of a compare method. Different problem.
Here is the general contract:
- If a<b and b<c, then a<c.
- If a==b, then compare(a, b) must return 0.
- If a<b then b > a.
- you may throw exceptions if you want.
You've broken the contract: If I pass the same value to your compare method, the final case occurs, and -1 is returned, which is saying: 'a is below a', and that breaks the rule.
Expand that last node (the -1) to: (a[1] < b[1]) ? -1 : 0.
答案2
得分: 0
a[1] 和 b[1] 交换时没有对称性。以下内容可以解决这个问题。
return (a[0] > b[0]) ? 1
: (a[0] < b[0]) ? -1
: (a[1] > b[1]) ? 1
: (a[1] < b[1]) ? -1
: 0;
对于这种结构性比较，应使用 Comparator 类。类似这样：
Arrays.sort(intervals, Comparator.comparing(c -> c[0]).thenComparing(c -> c[1]));
英文:
There is no symmetry when a[1] and b[1] are switched. The following would solve that.
return a[0] > b[0]) ? 1
: a[0] < b[0] ? -1
: a[1] > b[1] ? 1
: a[1] < b[1] ? -1
: 0;
The Comparator class should be used for such structural comparisons.
Something like:
Arrays.sort(intervals, Comparator.comparing(c -> c[0]).thenComparing(c -> c[1]));
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