探秘Collections.sort

摘要: 研究研究 JDK 的 sort 源码

先来随便写个

List<String> list = new ArrayList<String>();
Collections.sort(list);

里面是这样

@SuppressWarnings("unchecked")
public static <T extends Comparable<? super T>> void sort(List<T> list) {
list.sort(null);
}

list 的泛型需要继承 Comparable 接口，否则需要使用

@SuppressWarnings({"unchecked", "rawtypes"})
public static <T> void sort(List<T> list, Comparator<? super T> c) {
list.sort(c);
}

list.sort 长这样

@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);
}
}

1，转成数组；2，调用 Arrays.sort 排序；3，通过迭代器赋值回原来集合。

然后关键来了，JDK 默认的的 sort 方法到底有多神奇呢？

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);
}

再继续扒皮...

private static void legacyMergeSort(Object[] a) {
Object[] aux = a.clone();
mergeSort(aux, a, 0, a.length, 0);
}

这里用了一个 clone() 方法，mergeSort 看来已经是归并排序了，是不是一会儿还得赋值回来？继续看...

@SuppressWarnings({"unchecked", "rawtypes"})
private static void mergeSort(Object[] src,
Object[] dest,
int low,
int high,
int off) {
int length = high - low;

//这里这个常量是7，长度小于7直接冒泡解决，比较的是 dest 数组
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;
//从上面看下来off 是0，这里可以忽略
low  += off;
high += off;
//新技能 Get
int mid = (low + high) >>> 1;
mergeSort(dest, src, low, mid, -off);
mergeSort(dest, src, mid, high, -off);

// 如果列表已经有序, 只需要从 src 拷贝到 dest. 下面是优化过的拷贝算法
if (((Comparable)src[mid-1]).compareTo(src[mid]) <= 0) {
System.arraycopy(src, low, dest, destLow, length);
return;
}

// 归并算法的最后，拷贝到 dest
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++];
}
}

等等，好像什么地方漏了。。。。

TimSort.sort(a, 0, a.length, c, null, 0, 0);

看方法上的注释说这是自1.8开始后才使用的，尽可能使用给定的数据空间，为了提升性能而设计

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

// 如果数组长度小(这里是32), 使用0拷贝的"mini-TimSort"方式
if (nRemaining < MIN_MERGE) {
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;
}

擦，还得继续扒皮 binarySort

@SuppressWarnings("fallthrough")
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).
*/
while (left < right) {
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
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;
}
}

这是一个二分的插入排序算法。然后研究了下下面这个函数，返回连续升序数据的数据量，如果前 n 个数据是连续降序排列的，翻转此序列，并返回 n 。为了防止最坏的情况

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
while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0)
runHi++;
reverseRange(a, lo, runHi);
} else {                              // Ascending
while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0)
runHi++;
}

return runHi - lo;
}

TimeSort的逻辑还是挺复杂，这里引用一个 demo 的解释

http://blog.sina.com.cn/s/blog_8e6f1b330101h7fa.html

5. Demo
这一节用一个具体的例子来演示整个算法的演进过程：
*注意*：为了演示方便，我将TimSort中的minRun直接设置为2，否则我不能用很小的数组演示。。。同时把MIN_MERGE也改成2（默认为32），这样避免直接进入binary sort。

初始数组为[7,5,1,2,6,8,10,12,4,3,9,11,13,15,16,14]
=> 寻找连续的降序或升序序列 (4.3.2)
[1,5,7] [2,6,8,10,12,4,3,9,11,13,15,16,14]
=> 入栈 (4.3.4)
当前的栈区块为[3]
=> 进入merge循环 (4.3.5)
do not merge因为栈大小仅为1
=> 寻找连续的降序或升序序列 (4.3.2)
[1,5,7] [2,6,8,10,12] [4,3,9,11,13,15,16,14]
=> 入栈 (4.3.4)
当前的栈区块为[3, 5]
=> 进入merge循环 (4.3.5)
merge因为runLen[0]<=runLen[1]
1) gallopRight：寻找run1的第一个元素应当插入run0中哪个位置（”2”应当插入”1”之后），然后就可以忽略之前run0的元素（都比run1的第一个元素小）
2) gallopLeft：寻找run0的最后一个元素应当插入run1中哪个位置（”7”应当插入”8”之前），然后就可以忽略之后run1的元素（都比run0的最后一个元素大）
这样需要排序的元素就仅剩下[5,7] [2,6]，然后进行mergeLow
完成之后的结果：
[1,2,5,6,7,8,10,12] [4,3,9,11,13,15,16,14]
=> 入栈 (4.3.4)
当前的栈区块为[8]
退出当前merge循环因为栈中的区块仅为1
=> 寻找连续的降序或升序序列 (4.3.2)
[1,2,5,6,7,8,10,12] [3,4] [9,11,13,15,16,14]
=> 入栈 (4.3.4)
当前的栈区块大小为[8,2]
=> 进入merge循环 (4.3.5)
do not merge因为runLen[0]>runLen[1]
=> 寻找连续的降序或升序序列 (4.3.2)
[1,2,5,6,7,8,10,12] [3,4] [9,11,13,15,16] [14]
=> 入栈 (4.3.4)
当前的栈区块为[8,2,5]
=>
do not merege run1与run2因为不满足runLen[0]<=runLen[1]+runLen[2]
merge run2与run3因为runLen[1]<=runLen[2]
1) gallopRight：发现run1和run2就已经排好序
完成之后的结果：
[1,2,5,6,7,8,10,12] [3,4,9,11,13,15,16] [14]
=> 入栈 (4.3.4)
当前入栈的区块大小为[8,7]
退出merge循环因为runLen[0]>runLen[1]
=> 寻找连续的降序或升序序列 (4.3.2)
最后只剩下[14]这个元素：[1,2,5,6,7,8,10,12] [3,4,9,11,13,15,16] [14]
=> 入栈 (4.3.4)
当前入栈的区块大小为[8,7,1]
=> 进入merge循环 (4.3.5)
merge因为runLen[0]<=runLen[1]+runLen[2]
因为runLen[0]>runLen[2]，所以将run1和run2先合并。（否则将run0和run1先合并）
1) gallopRight & 2) gallopLeft
这样需要排序的元素剩下[13,15] [14]，然后进行mergeHigh
完成之后的结果：
[1,2,5,6,7,8,10,12] [3,4,9,11,13,14,15,16] 当前入栈的区块为[8,8]
=>
继续merge因为runLen[0]<=runLen[1]
1) gallopRight & 2) gallopLeft
需要排序的元素剩下[5,6,7,8,10,12] [3,4,9,11]，然后进行mergeHigh
完成之后的结果：
[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16] 当前入栈的区块大小为[16]
=>
不需要final merge因为当前栈大小为1
=>
结束