JDK错误用法—TimSort

private void compositeRank(List<ADXEntity> list) {
   if (CollectionUtils.isEmpty(list)) {
       return;
    }
    list.sort((o1, o2) -> o1.getQ() > o2.getQ() ? -1 : 1);
}

@Override
@SuppressWarnings("unchecked")
public void sort(Comparator<? super E> c) {
    final int expectedModCount = modCount;
    Arrays.sort((E[]) elementData, 0, size, c);
    if (modCount != expectedModCount) {
     throw new ConcurrentModificationException();
    }
    modCount++;
    }
}

public static <T> void sort(T[] a, int fromIndex, int toIndex,Comparator<? super T> c) {
    if (c == null) {
        sort(a, fromIndex, toIndex);
    } else {
        rangeCheck(a.length, fromIndex, toIndex);
        if (LegacyMergeSort.userRequested)
            legacyMergeSort(a, fromIndex, toIndex, c);
        else
            TimSort.sort(a, fromIndex, toIndex, c, null, 0, 0);
    }
}

static <T> void sort(T[] a,int lo,int hi, Comparator<? super T> c,T[] work, int workBase, int workLen) {
    assert c != null && a != null && lo >= 0 && lo <= hi && hi <= a.length;
    
    int nRemaining  = hi - lo;
    if (nRemaining < 2)
        return;  // 长度小于2，必然是有序的

// 如果数组长度小于MIN_MERGE（默认为32），则使用没有merge的二分排序法。
if (nRemaining < MIN_MERGE) {
    int initRunLen = countRunAndMakeAscending(a, lo, hi, c);
    binarySort(a, lo, hi, lo + initRunLen, c);
    return;
}
TimSort<T> ts = new TimSort<>(a, c, work, workBase, workLen);
//获取最短的run
int minRun = minRunLength(nRemaining);
do {
    // 找出连续升序的最大个数
    int runLen = countRunAndMakeAscending(a, lo, hi, c);

    // 如果太短，就通过二分插入法扩展
    if (runLen < minRun) {
        int force = nRemaining <= minRun ? nRemaining : minRun;
        binarySort(a, lo, lo + force, lo + runLen, c);
        runLen = force;
    }
    ts.pushRun(lo, runLen);// 将run入栈
    ts.mergeCollapse();// 合并

    // 继续寻找下一个run
    lo += runLen;
    nRemaining -= runLen;
} while (nRemaining != 0);

// 合并剩下的run
assert lo == hi;
ts.mergeForceCollapse();
assert ts.stackSize == 1;
}

private void mergeCollapse() {
    while (stackSize > 1) {
        int n = stackSize - 2;
        if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) {
            if (runLen[n - 1] < runLen[n + 1])
                n--;
            mergeAt(n);
        } else if (runLen[n] <= runLen[n + 1]) {
            mergeAt(n);
        } else {
            break; // Invariant is established
        }
    }
}

private void mergeAt(int i) {、
        assert stackSize >= 2;
        assert i >= 0;
        assert i == stackSize - 2 || i == stackSize - 3;

        int base1 = runBase[i];
        int len1 = runLen[i];
        int base2 = runBase[i + 1];
        int len2 = runLen[i + 1];
        assert len1 > 0 && len2 > 0;
        assert base1 + len1 == base2;

        /*
         * 记录归并这些run的长度；如果i是最后一个run的三
         * 个run中第一个，就滑过最后一个run（最后一个
         *run不会被归并），而i+1在任何情况都会消失。
         * 其实这是与mergeCollapse的逻辑相关联的，传入          *的参数i，已经明确了它必须是被归并的，而与它相关          *联的就只有i+1
         */
        runLen[i] = len1 + len2;
        if (i == stackSize - 3) {//如果传入的是3个数中的第一个
            runBase[i + 1] = runBase[i + 2];
            runLen[i + 1] = runLen[i + 2];
        }
        stackSize--;

        /*
         * 查找run2的第一个元素在run1中的位置。可以忽略
         *run1中的先前元素(因为它们已经就绪)。
         */
        int k = gallopRight(a[base2], a, base1, len1, 0, c);
        assert k >= 0;
        base1 += k;
        len1 -= k;
        if (len1 == 0)
            return;

        /*
         * 查找run1的最后一个元素在run2中的位置。
         * run2后续的元素可以被忽略了，因为他们
         * 本来就是有序的。
         */
        len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c);
        assert len2 >= 0;
        if (len2 == 0)
            return;

        // 归并排序，使用min(len1,len2)作为临时数组
        if (len1 <= len2)
            mergeLo(base1, len1, base2, len2);
        else
            mergeHi(base1, len1, base2, len2);
    }

private void mergeLo(int base1, int len1, int base2, int len2) {
        assert len1 > 0 && len2 > 0 && base1 + len1 == base2;

        // Copy first run into temp array
        T[] a = this.a; // For performance
        T[] tmp = ensureCapacity(len1);
        int cursor1 = tmpBase; // Indexes into tmp array
        int cursor2 = base2;   // Indexes int a
        int dest = base1;      // Indexes int a
        System.arraycopy(a, base1, tmp, cursor1, len1);

       //将第二个run中的第一个num移动到，整个第一个run的第一个num的位置上，因为从前面的gallopLeft结果也可知道，a[cusor2]相对于a[dest]是递增的；
        a[dest++] = a[cursor2++];
        if (--len2 == 0) {
            System.arraycopy(tmp, cursor1, a, dest, len1);
            return;
        }
        if (len1 == 1) {
            System.arraycopy(a, cursor2, a, dest, len2);
            a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
            return;
        }

        Comparator<? super T> c = this.c;  // Use local variable for performance
        int minGallop = this.minGallop;
        
    outer:
        while (true) {
            int count1 = 0; // Number of times in a row that first run won
            int count2 = 0; // Number of times in a row that second run won

            /*
             * Do the straightforward thing until (if ever) one run starts
             * winning consistently.
             */
            do {
                assert len1 > 1 && len2 > 0;
                if (c.compare(a[cursor2], tmp[cursor1]) < 0) {
                    a[dest++] = a[cursor2++];
                    count2++;
                    count1 = 0;
                    if (--len2 == 0)
                        break outer;
                } else {
                    a[dest++] = tmp[cursor1++];
                    count1++;
                    count2 = 0;
                    if (--len1 == 1)
                        break outer;
                }
            } while ((count1 | count2) < minGallop);

            /*
             * One run is winning so consistently that galloping may be a
             * huge win. So try that, and continue galloping until (if ever)
             * neither run appears to be winning consistently anymore.
             */
            do {
                assert len1 > 1 && len2 > 0;
                count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c);
                if (count1 != 0) {
                    System.arraycopy(tmp, cursor1, a, dest, count1);
                    dest += count1;
                    cursor1 += count1;
                    len1 -= count1;
                    if (len1 <= 1) // len1 == 1 || len1 == 0
                        break outer;
                }
                a[dest++] = a[cursor2++];
                if (--len2 == 0)
                    break outer;

                count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c);
                if (count2 != 0) {
                    System.arraycopy(a, cursor2, a, dest, count2);
                    dest += count2;
                    cursor2 += count2;
                    len2 -= count2;
                    if (len2 == 0)
                        break outer;
                }
                a[dest++] = tmp[cursor1++];
                if (--len1 == 1)
                    break outer;
                minGallop--;
            } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
            if (minGallop < 0)
                minGallop = 0;
            minGallop += 2;  // Penalize for leaving gallop mode
        }  // End of "outer" loop
        this.minGallop = minGallop < 1 ? 1 : minGallop;  // Write back to field

        if (len1 == 1) {
            assert len2 > 0;
            System.arraycopy(a, cursor2, a, dest, len2);
            a[dest + len2] = tmp[cursor1]; //  Last elt of run 1 to end of merge
        } else if (len1 == 0) {
            throw new IllegalArgumentException(
                "Comparison method violates its general contract!");
        } else {
            assert len2 == 0;
            assert len1 > 1;
            System.arraycopy(tmp, cursor1, a, dest, len1);
        }
    }