概要
這個類在 Oracle 的官方文檔裏是查不到的,但是確實在 OpenJDK
的源代碼裏出現了,Arrays
中的 sort
函數用到了這個用於排序的類。它將歸併排序(merge
sort) 與插入排序(insertion sort) 結合,並進行了一些優化。對於已經部分排序的數組,時間複雜度遠低於 O(n log(n))
,最好可達 O(n)
,對於隨機排序的數組,時間複雜度爲 O(nlog(n))
,平均時間複雜度 O(nlog(n))
。強烈建議在看此文前觀看
Youtube 上的 可視化Timsort,看完後馬上就會對算法的執行過程有一個感性的瞭解。然後,可以閱讀 Wikipeida 詞條:Timsort。
這個排序算法在 Java SE 7, Android, GNU Octave 中都得到了應用。另外,文 後也推薦了兩篇非常好的文章,如果想搞明白 TimSort
最好閱讀一下。
此類是對 Python
中,由 Tim
Peters
實現的排序算法的改寫。實現來自:listobject.c.
原始論文來自:
"Optimistic Sorting and Information Theoretic Complexity" Peter
McIlroy SODA (Fourth Annual ACM-SIAM Symposium on Discrete
Algorithms), pp 467-474, Austin, Texas, 25-27 January 1993.
實現
- sort
static <T> void sort(T[] a, int lo, int hi, Comparator<? super T> c) {
if (c == null) {
Arrays.sort(a, lo, hi);
return;
}
rangeCheck(a.length, lo, hi);
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) {
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);
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;
}
下面分段解釋:
if (c == null) {
Arrays.sort(a, lo, hi);
return;
}
如果沒有提供 Comparaotr
的話,會調用 Arrays.sort
中的函數,背後其實又會調用 ComparableTimSort
,它是對沒有提供Comparator
,但是實現了 Comparable
的元素進行排序,算法和這裏的是一樣的,就是元素比較方法不一樣。
在後來 jdk8 的實現中,此段代碼被替換爲
assert c != null && a != null && lo >= 0 && lo <= hi && hi <= a.length;
對輸入有了更嚴格的要求。
後面是算法的主體:
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) {
int initRunLen = countRunAndMakeAscending(a, lo, hi, c);
binarySort(a, lo, hi, lo + initRunLen, c);
return;
}
- 如果元素個數小於2,直接返回,因爲這兩個元素已經排序了
- 如果元素個數小於一個閾值(默認爲 32,這是個經驗值,Python 實現裏是 64,需要是 2^n),調用
binarySort
,這是一個不包含合併操作的mini-TimSort
。 - 在關鍵的
do-while
循環中,不斷地進行排序,合併,排序,合併,一直到所有數據都處理完。
TimSort<T> ts = new TimSort<>(a, c);
int minRun = minRunLength(nRemaining);
do {
...
} while (nRemaining != 0);
- minRunLength
這個函數會找出 run
的最小長度,少於這個長度就需要對其進行擴展。
static int minRunLength(int n) {
assert n >= 0;
int r = 0; // Becomes 1 if any 1 bits are shifted off
while (n >= MIN_MERGE) {
r |= (n & 1);
n >>= 1;
}
return n + r;
}
先看看 n 與 minRunLength(n) 對應關係
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10
11 11
12 12
13 13
14 14
15 15
16 16
17 17
18 18
19 19
20 20
21 21
22 22
23 23
24 24
25 25
26 26
27 27
28 28
29 29
30 30
31 31
32 16
33 17
34 17
35 18
36 18
37 19
38 19
39 20
40 20
41 21
42 21
43 22
44 22
45 23
46 23
47 24
48 24
49 25
50 25
51 26
52 26
53 27
54 27
55 28
56 28
57 29
58 29
59 30
60 30
61 31
62 31
63 32
64 16
65 17
66 17
67 17
68 17
69 18
70 18
71 18
72 18
73 19
74 19
75 19
76 19
77 20
78 20
79 20
80 20
81 21
82 21
83 21
84 21
85 22
86 22
87 22
88 22
89 23
90 23
91 23
92 23
93 24
94 24
95 24
96 24
97 25
98 25
99 25
...
看這個估計可以猜出來函數的功能了,下面解釋一下。
這個函數根據 n 計算出對應的 natural run
的最小長度。MIN_MERGE
默認爲 32
,如果n小於此值,那麼返回 n
本身。否則會將 n
不斷地右移,直到少於 MIN_MERGE
,同時記錄一個 r
值,r
代表最後一次移位n時,n最低位是0還是1。 最後返回 n + r
,這也意味着只保留最高的 5 位,再加上第六位。
- do-while
我們再看看 do-while
中發生了什麼。
TimSort<T> ts = new TimSort<>(a, c);
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);
countRunAndMakeAscending
會找到一個 run
,這個 run
必須是已經排序的,並且函數會保證它爲升序,也就是說,如果找到的是一個降序的,會對其進行翻轉。
簡單看一眼這個函數:
- countRunAndMakeAscending
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;
}
注意其中的 reverseRange
就是我們說的翻轉。
現在,有必要看一下 binarySort
了。
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;
}
}
我們都聽說過 binarySearch
,但是這個 binarySort
又是什麼呢? binarySort
對數組 a[lo:hi]
進行排序,並且a[lo:start]
是已經排好序的。算法的思路是對 a[start:hi]
中的元素,每次使用 binarySearch
爲它在 a[lo:start]
中找到相應位置,並插入。
回到 do-while
循環中,看看 binarySearch
的作用:
// 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;
}
所以,我們明白了,binarySort
對 run
進行了擴展,並且擴展後,run
仍然是有序的。
隨後:
// Push run onto pending-run stack, and maybe merge
ts.pushRun(lo, runLen);
ts.mergeCollapse();
// Advance to find next run
lo += runLen;
nRemaining -= runLen;
當前的 run
位於 a[lo:runLen]
,將其入棧,然後將棧中的 run
合併。
- pushRun
private void pushRun(int runBase, int runLen) {
this.runBase[stackSize] = runBase;
this.runLen[stackSize] = runLen;
stackSize++;
}
入棧過程簡單明瞭,不解釋。
再看另一個關鍵函數,合併操作。如果你看過文章開頭提到的對 Timsort
進行可視化的視頻,一定會對合並操作印象深刻。它會把已經排序的 run
合併成一個大 run
,此大 run
也會排好序。
/**
* Examines the stack of runs waiting to be merged and merges adjacent runs
* until the stack invariants are reestablished:
*
* 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1]
* 2. runLen[i - 2] > runLen[i - 1]
*
* This method is called each time a new run is pushed onto the stack,
* so the invariants are guaranteed to hold for i < stackSize upon
* entry to the method.
*/
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
}
}
}
合併的過程會一直循環下去,一直到註釋裏提到的循環不變式得到滿足。
- mergeAt
mergeAt
會把棧頂的兩個 run
合併起來:
/**
* Merges the two runs at stack indices i and i+1. Run i must be
* the penultimate or antepenultimate run on the stack. In other words,
* i must be equal to stackSize-2 or stackSize-3.
*
* @param i stack index of the first of the two runs to merge
*/
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;
/*
* Record the length of the combined runs; if i is the 3rd-last
* run now, also slide over the last run (which isn't involved
* in this merge). The current run (i+1) goes away in any case.
*/
runLen[i] = len1 + len2;
if (i == stackSize - 3) {
runBase[i + 1] = runBase[i + 2];
runLen[i + 1] = runLen[i + 2];
}
stackSize--;
/*
* Find where the first element of run2 goes in run1. Prior elements
* in run1 can be ignored (because they're already in place).
*/
int k = gallopRight(a[base2], a, base1, len1, 0, c);
assert k >= 0;
base1 += k;
len1 -= k;
if (len1 == 0)
return;
/*
* Find where the last element of run1 goes in run2. Subsequent elements
* in run2 can be ignored (because they're already in place).
*/
len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c);
assert len2 >= 0;
if (len2 == 0)
return;
// Merge remaining runs, using tmp array with min(len1, len2) elements
if (len1 <= len2)
mergeLo(base1, len1, base2, len2);
else
mergeHi(base1, len1, base2, len2);
}
由於要合併的兩個 run
是已經排序的,所以合併的時候,有會特別的技巧。假設兩個 run
是 run1,run2
,先用 gallopRight
在 run1
裏使用 binarySearch
查找 run2
首元素
的位置 k
, 那麼 run1
中 k
前面的元素就是合併後最小的那些元素。然後,在 run2
中查找 run1
尾元素
的位置 len2
,那麼 run2
中 len2
後面的那些元素就是合併後最大的那些元素。最後,根據len1
與 len2
大小,調用 mergeLo
或者 mergeHi
將剩餘元素合併。
gallop
和 merge
就不展開了。
另外,強烈推薦閱讀文後的兩篇文章,第一篇可以看到 JDK7 中更換排序算法後可能引發的問題,另外,也會介紹源代碼,並給出具體的例子。第二篇會告訴你如何對一個 MergeSort
進行優化,介紹了 TimSort
背後的思想。
如果對代碼有更多見解,可以寫在 rtfcode-java-1.8-TimSort
推薦閱讀