Caused by: java.lang.IllegalArgumentException: Comparison method violates its general contract!
问题
导致该异常的原因有以下两点构成：1.使用了jdk7 2.使用了比较器，并且比较器违反了比较规则
以下是关于比较器不兼容的声明：
Area: API: Utilities
Synopsis: Updated sort behavior for Arrays and Collections may throw an IllegalArgumentException
Description: The sorting algorithm used by java.util.Arrays.sort and (indirectly) by java.util.Collections.sort has been replaced.
The new sort implementation may throw an IllegalArgumentException if it detects a Comparable that violates the Comparable contract.
The previous implementation silently ignored such a situation.
If the previous behavior is desired, you can use the new system property, java.util.Arrays.useLegacyMergeSort,
to restore previous mergesort behavior.
Nature of Incompatibility: behavioral
RFE: 6804124
描述的意思是说，java.util.Arrays.sort(java.util.Collections.sort调用的也是此方法)方法中的排序算法在JDK7中已经被替换了。如果违法了比较的约束新的排序算法也许会抛出llegalArgumentException异常。JDK6中的实现则忽略了这种情况。那么比较的约束是什么呢？看这里，大体如下：
sgn(compare(x, y)) == -sgn(compare(y, x))
((compare(x, y)>0) && (compare(y, z)>0)) implies compare(x, z)>0
compare(x, y)==0 implies that sgn(compare(x, z))==sgn(compare(y, z)) for all z
当x == y时，sgn(compare(x, y)) = -1，-sgn(compare(y, x)) = 1，这违背了sgn(compare(x, y)) == -sgn(compare(y, x))约束，所以在JDK7中抛出了本文标题的异常。
解决方式
1.JVM加入如下参数-Djava.util.Arrays.useLegacyMergeSort=true，表示使用JDK6的排序算法
2.按照规定的比较规则进行值的返回，a==b 返回 0，ab 返回 1
