ArrayList，排序方法的调用过程
// 排序方法
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) {
// 如果没有实现比较方法
if (c == null) {
sort(a, fromIndex, toIndex);
} else {
rangeCheck(a.length, fromIndex, toIndex);
if (Arrays.LegacyMergeSort.userRequested)
legacyMergeSort(a, fromIndex, toIndex, c);
else
TimSort.sort(a, fromIndex, toIndex, c, null, 0, 0);
}
} public static void sort(Object[] a, int fromIndex, int toIndex) {
rangeCheck(a.length, fromIndex, toIndex);
//经查资料，这是个传统的归并排序，需要通过设置系统属性后，才能进行调用
// System.setProperty("java.util.Arrays.useLegacyMergeSort", "true");
if (Arrays.LegacyMergeSort.userRequested)
legacyMergeSort(a, fromIndex, toIndex);
else
ComparableTimSort.sort(a, fromIndex, toIndex, null, 0, 0);
}
然后继续看下在没有实现Comparator接口的情况，传统归并排序的实现
private static void legacyMergeSort(Object[] a,
int fromIndex, int toIndex) {
// 复制对应范围的数组
Object[] aux = copyOfRange(a, fromIndex, toIndex);
mergeSort(aux, a, fromIndex, toIndex, -fromIndex);
} // 使用插入排序进行优化的归并排序
// dest为要排序的数组
// off:负数，因为aux是复制a数组fromIndex-toIndex范围的数据，但位置从0开始，fromIndex-off则为aux开始的坐标
private static void mergeSort(Object[] src,
Object[] dest,
int low,
int high,
int off) {
int length = high - low; // Insertion sort on smallest arrays
// 长度小于7，则使用插入排序
if (length < INSERTIONSORT_THRESHOLD) {
for (int i=low; i<high; i++)
for (int j=i; j>low &&
((Comparable) dest[j-1]).compareTo(dest[j])>0; j--)
swap(dest, j, j-1);
return;
} // Recursively sort halves of dest into src
int destLow = low;
int destHigh = high;
low += off;
high += off;
int mid = (low + high) >>> 1; // 交换dest,src位置，这样就能排序src，然后合并到dest
// 注意这个递归的开始，dest为要排序的数组a，所以最终a会合并成有序的数组
mergeSort(dest, src, low, mid, -off);
mergeSort(dest, src, mid, high, -off); // If list is already sorted, just copy from src to dest. This is an
// optimization that results in faster sorts for nearly ordered lists.
// 如果说src排完序后，两个范围的src，刚好前一个小于后一个，则直接复制。
// 例如两个范围【1，2】【3，4】，其实他们已经有序【1，2，3，4】
if (((Comparable)src[mid-1]).compareTo(src[mid]) <= 0) {
System.arraycopy(src, low, dest, destLow, length);
return;
} // Merge sorted halves (now in src) into dest
// 可以看成两个有序的数组之间进行合并
for(int i = destLow, p = low, q = mid; i < destHigh; i++) {
if (q >= high || p < mid && ((Comparable)src[p]).compareTo(src[q])<=0)
dest[i] = src[p++];
else
dest[i] = src[q++];
}
} /**
* Swaps x[a] with x[b].
*/
private static void swap(Object[] x, int a, int b) {
Object t = x[a];
x[a] = x[b];
x[b] = t;
}
再看看当前默认使用的排序方法（没使用Comparable的情况）
/**
*
* @param a 待排序的数组
* @param lo 开始位置，包括当前位置
* @param hi 结束位置，不包括当前位置
* @param work
* @param workBase
* @param workLen
*/
static void sort(Object[] a, int lo, int hi, Object[] work, int workBase, int workLen) {
assert a != null && lo >= 0 && lo <= hi && hi <= a.length; int nRemaining = hi - lo;
if (nRemaining < 2)
return; // Arrays of size 0 and 1 are always sorted // If array is small, do a "mini-TimSort" with no merges
// 排序的范围小于32的情况
if (nRemaining < MIN_MERGE) {
// 从lo位置开始，返回最长的递增序列长度，（下面有介绍）
int initRunLen = countRunAndMakeAscending(a, lo, hi);
// 折半插入排序，（下面有介绍）
binarySort(a, lo, hi, lo + initRunLen);
return;
} /**
* March over the array once, left to right, finding natural runs,
* extending short natural runs to minRun elements, and merging runs
* to maintain stack invariant.
*/
ComparableTimSort ts = new ComparableTimSort(a, work, workBase, workLen);
int minRun = minRunLength(nRemaining);
do {
// Identify next run
int runLen = countRunAndMakeAscending(a, lo, hi); // If run is short, extend to min(minRun, nRemaining)
if (runLen < minRun) {
int force = nRemaining <= minRun ? nRemaining : minRun;
binarySort(a, lo, lo + force, lo + runLen);
runLen = force;
} // Push run onto pending-run stack, and maybe merge
ts.pushRun(lo, runLen);
ts.mergeCollapse(); // Advance to find next run
lo += runLen;
nRemaining -= runLen;
} while (nRemaining != 0); // Merge all remaining runs to complete sort
assert lo == hi;
ts.mergeForceCollapse();
assert ts.stackSize == 1;
}
具体看看里面的方法实现
/**
* 返回从lo开始最长递增序列
* @param a 待排序的数组
* @param lo 开始位置，包括当前位置
* @param hi 结束位置，不包含当前位置
* @return
*/
private static int countRunAndMakeAscending(Object[] a, int lo, int hi) {
assert lo < hi;
int runHi = lo + 1;
if (runHi == hi)
return 1; // Find end of run, and reverse range if descending
// 如果lo+1位置的数小于lo位置的数，就找出最长的递减序列，然后进行反转
if (((Comparable) a[runHi++]).compareTo(a[lo]) < 0) { // Descending
while (runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) < 0)
runHi++;
reverseRange(a, lo, runHi);
} else { // Ascending
while (runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) >= 0)
runHi++;
} // 返回递增的长度
return runHi - lo;
} // 反转
private static void reverseRange(Object[] a, int lo, int hi) {
hi--;
while (lo < hi) {
Object t = a[lo];
a[lo++] = a[hi];
a[hi--] = t;
}
} /**
* 折半插入排序
* @param a 待排序的数组
* @param lo 开始位置，包括当前位置
* @param hi 结束位置，不包含当前位置
* @param start start以前为递增序列
*/
private static void binarySort(Object[] a, int lo, int hi, int start) {
assert lo <= start && start <= hi;
if (start == lo)
start++;
// 从start开始遍历
for ( ; start < hi; start++) {
Comparable pivot = (Comparable) a[start]; // Set left (and right) to the index where a[start] (pivot) belongs
int left = lo;
int right = start;
assert left <= right;
/*
* Invariants:
* pivot >= all in [lo, left).
* pivot < all in [right, start).
*/
// 二分找lo到start（不包括start）范围内，找到pivot适合插入的位置
while (left < right) {
int mid = (left + right) >>> 1;
if (pivot.compareTo(a[mid]) < 0)
right = mid;
else
left = mid + 1;
}
assert left == right; /*
* The invariants still hold: pivot >= all in [lo, left) and
* pivot < all in [left, start), so pivot belongs at left. Note
* that if there are elements equal to pivot, left points to the
* first slot after them -- that's why this sort is stable.
* Slide elements over to make room for pivot.
*/
int n = start - left; // The number of elements to move
// Switch is just an optimization for arraycopy in default case
// 如果只需后移2位或者1位，直接换值
// 更多，则调用系统方法进行复制
switch (n) {
case 2: a[left + 2] = a[left + 1];
case 1: a[left + 1] = a[left];
break;
default: System.arraycopy(a, left, a, left + 1, n);
}
// 后移成功后，赋值
a[left] = pivot;
}
}
没分析完。。。需要学习一波tim sort。。