最近看了一些排序相关的文章，因此比较好奇，Java中的排序是如何做的。本片文章介绍的是JDK1.8，List中的sort方法。
先来看看List中的sort是怎么写的：
@SuppressWarnings({"unchecked", "rawtypes"})
default void sort(Comparator<? super E> c) {
Object[] a = this.toArray();
Arrays.sort(a, (Comparator) c);
ListIterator<E> i = this.listIterator();
for (Object e : a) {
i.next();
i.set((E) e);
}
}
首先，你需要传入一个比较器作为参数，这个好理解，毕竟你肯定要定一个比较标准。然后就是将list转换成一个数组，再对这个数组进行排序，排序完之后，再利用iterator重新改变list。
接着，我们再来看看Arrays.sort：
public static <T> void sort(T[] a, Comparator<? super T> c) {
if (c == null) {
sort(a);
} else {
if (LegacyMergeSort.userRequested)
legacyMergeSort(a, c);
else
TimSort.sort(a, 0, a.length, c, null, 0, 0);
}
}
public static void sort(Object[] a) {
if (LegacyMergeSort.userRequested)
legacyMergeSort(a);
else
ComparableTimSort.sort(a, 0, a.length, null, 0, 0);
}
static final class LegacyMergeSort {
private static final boolean userRequested =
java.security.AccessController.doPrivileged(
new sun.security.action.GetBooleanAction(
"java.util.Arrays.useLegacyMergeSort")).booleanValue();
}
这样可以看出，其实排序的核心就是TimSort，LegacyMergeSort大致意思是表明如果版本很旧的话，就用这个，新版本是不会采用这种排序方式的。
我们再来看看TimSort的实现：
private static final int MIN_MERGE = 32;
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; // Arrays of size 0 and 1 are always sorted
// If array is small, do a "mini-TimSort" with no merges
if (nRemaining < MIN_MERGE) {
// 获得最长的递增序列
int initRunLen = countRunAndMakeAscending(a, lo, hi, c);
binarySort(a, lo, hi, lo + initRunLen, c);
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.
*/
TimSort<T> ts = new TimSort<>(a, c, work, workBase, workLen);
int minRun = minRunLength(nRemaining);
do {
// Identify next run
int runLen = countRunAndMakeAscending(a, lo, hi, c);
// 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, c);
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;
}
如果小于2个，代表不再不需要排序；如果小于32个，则采用优化的二分排序。怎么优化的呢？首先获得最长的递增序列：
private static <T> int countRunAndMakeAscending(T[] a, int lo, int hi,
Comparator<? super T> c) {
assert lo < hi;
int runHi = lo + 1;
if (runHi == hi)
return 1;
// Find end of run, and reverse range if descending
if (c.compare(a[runHi++], a[lo]) < 0) { // Descending
// 一开始是递减序列，就找出最长递减序列的最后一个下标
while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0)
runHi++;
// 逆转前面的递减序列
reverseRange(a, lo, runHi);
} else { // Ascending
while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0)
runHi++;
}
return runHi - lo;
}
接着进行二分排序：
private static <T> void binarySort(T[] a, int lo, int hi, int start,
Comparator<? super T> c) {
assert lo <= start && start <= hi;
if (start == lo)
start++;
for ( ; start < hi; start++) {
T pivot = 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).
*/
// start位置是递增序列后的第一个数的位置
// 从前面的递增序列中找出start位置的数应该处于的位置
while (left < right) {
// >>> 无符号右移
int mid = (left + right) >>> 1;
if (c.compare(pivot, 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
// 比pivot大的数往后移动一位
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);
}
a[left] = pivot;
}
}
好了，待排序数量小于32个的讲完了，现在来说说大于等于32个情况。首先，获得一个叫minRun的东西，这是个啥含义呢：
int minRun = minRunLength(nRemaining);
private static int minRunLength(int n) {
assert n >= 0;
int r = 0; // Becomes 1 if any 1 bits are shifted off
while (n >= MIN_MERGE) {
// 这里我没搞懂的是为什么不直接将(n & 1)赋值给r，而要做一次逻辑或。
r |= (n & 1);
n >>= 1;
}
return n + r;
}
各种位运算符，MIN_MERGE默认为32，如果n小于此值，那么返回n本身。否则会将n不断地右移，直到小于MIN_MERGE，同时记录一个r值，r代表最后一次移位n时，n最低位是0还是1。
其实看注释比较容易理解：
Returns the minimum acceptable run length for an array of the specified length. Natural runs shorter than this will be extended with binarySort.
Roughly speaking, the computation is: If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).
Else if n is an exact power of 2, return MIN_MERGE/2.
Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k is close to, but strictly less than, an exact power of 2. For the rationale, see listsort.txt.
返回结果其实就是用于接下来的合并排序中。
接下来就是一个while循环
do {
// Identify next run
// 获得一个最长递增序列
int runLen = countRunAndMakeAscending(a, lo, hi, c);
// If run is short, extend to min(minRun, nRemaining)
// 如果最长递增序列
if (runLen < minRun) {
int force = nRemaining <= minRun ? nRemaining : minRun;
binarySort(a, lo, lo + force, lo + runLen, c);
runLen = force;
}
// Push run onto pending-run stack, and maybe merge
// lo——runLen为将要被归并的范围
ts.pushRun(lo, runLen);
// 归并
ts.mergeCollapse();
// Advance to find next run
lo += runLen;
nRemaining -= runLen;
} while (nRemaining != 0);
这样，假设你的每次归并排序的两个序列为r1和r2，r1肯定是有序的，r2也已经被排成递增序列了，因此这样的归并排序就比较特殊了。
为什么要用归并排序呢，因为归并排序的时间复杂度永远为O(nlogn)，空间复杂度为O(n)，以空间换取时间。
总结
好了，以上就是针对Java中的排序做的一次总结，但具体的归并代码还没有分析，其实我自己也没有完全研究透，为什么minRun的取值是这样的，这也和TimSort中的stackLen有关，有兴趣的小伙伴可以在下方留言，我们可以一起探讨。
随时随地看视频
