OpenJDK 源代码阅读之 TimSort

概要

这个类在 Oracle 的官方文档里是查不到的,但是确实在 OpenJDK 的源代码里出现了,Arrays 中的 sort 函数用到了这个用于排序的类。它将归并排序(merge sort) 与插入排序(insertion sort) 结合,并进行了一些优化。对于已经部分排序的数组,时间复杂度远低于 O(n log(n)),最好可达 O(n),对于随机排序的数组,时间复杂度为 O(nlog(n)),平均时间复杂度 O(nlog(n))。强烈建议在看此文前观看 Youtube 上的 可视化Timsort,看完后马上就会对算法的执行过程有一个感性的了解。然后,可以阅读 Wikipeida 词条:Timsort。 这个排序算法在 Java SE 7, Android, GNU Octave 中都得到了应用。另外,文 后也推荐了两篇非常好的文章,如果想搞明白 TimSort 最好阅读一下。

此类是对 Python 中,由 Tim Peters 实现的排序算法的改写。实现来自:listobject.c.

原始论文来自:

"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.

实现

  • 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;
}

下面分段解释:

if (c == null) {
      
    Arrays.sort(a, lo, hi);
    return;
}

如果没有提供 Comparaotr 的话,会调用 Arrays.sort 中的函数,背后其实又会调用 ComparableTimSort,它是对没有提供Comparator ,但是实现了 Comparable 的元素进行排序,算法和这里的是一样的,就是元素比较方法不一样。

后面是算法的主体:

    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;
    }

  1. 如果元素个数小于2,直接返回,因为这两个元素已经排序了
  2. 如果元素个数小于一个阈值(默认为),调用 binarySort,这是一个不包含合并操作的 mini-TimSort
  3. 在关键的 do-while 循环中,不断地进行排序,合并,排序,合并,一直到所有数据都处理完。
    TimSort<T> ts = new TimSort<>(a, c);
    int minRun = minRunLength(nRemaining);
    do {
      

        ...

    } while (nRemaining != 0);
  • minRunLength

这个函数会找出 run 的最小长度,少于这个长度就需要对其进行扩展。

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 与 minRunLength(n) 对应关系

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

...

看这个估计可以猜出来函数的功能了,下面解释一下。

这个函数根据 n 计算出对应的 natural run 的最小长度。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,直到少于 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。 最后返回 n + r,这也意味着只保留最高的 5 位,再加上第六位。

  • do-while

我们再看看 do-while 中发生了什么。

   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 会找到一个 run ,这个 run 必须是已经排序的,并且函数会保证它为升序,也就是说,如果找到的是一个降序的,会对其进行翻转。

简单看一眼这个函数:

  • 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 就是我们说的翻转。

现在,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;
    }
}

我们都听说过 binarySearch ,但是这个 binarySort 又是什么呢? binarySort 对数组 a[lo:hi] 进行排序,并且a[lo:start] 是已经排好序的。算法的思路是对 a[start:hi] 中的元素,每次使用 binarySearch 为它在 a[lo:start] 中找到相应位置,并插入。

回到 do-while 循环中,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,我们明白了,binarySort 对 run 进行了扩展,并且扩展后,run 仍然是有序的。

随后:

   // Push run onto pending-run stack, and maybe merge
    ts.pushRun(lo, runLen);
    ts.mergeCollapse();

    // Advance to find next run
    lo += runLen;
    nRemaining -= runLen;

当前的 run 位于 a[lo:runLen] ,将其入栈,然后将栈中的 run 合并。

  • pushRun
private void pushRun(int runBase, int runLen) {
      
    this.runBase[stackSize] = runBase;
    this.runLen[stackSize] = runLen;
    stackSize++;
}

入栈过程简单明了,不解释。

再看另一个关键函数,合并操作。如果你看过文章开头提到的对 Timsort 进行可视化的视频,一定会对合并操作印象深刻。它会把已经排序的 run 合并成一个大 run,此大 run 也会排好序。

/**
 * 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
        }
    }
}

合并的过程会一直循环下去,一直到注释里提到的循环不变式得到满足。

  • mergeAt

mergeAt 会把栈顶的两个 run 合并起来:

   /**
     * 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);
}

由于要合并的两个 run 是已经排序的,所以合并的时候,有会特别的技巧。假设两个 run 是 run1,run2 ,先用 gallopRight在 run1 里使用 binarySearch 查找 run2 首元素 的位置 k, 那么 run1 中 k 前面的元素就是合并后最小的那些元素。然后,在 run2 中查找 run1 尾元素 的位置 len2 ,那么 run2 中 len2 后面的那些元素就是合并后最大的那些元素。最后,根据len1 与 len2 大小,调用 mergeLo 或者 mergeHi 将剩余元素合并。

gallop 和 merge 就不展开了。

另外,强烈推荐阅读文后的两篇文章,第一篇可以看到 JDK7 中更换排序算法后可能引发的问题,另外,也会介绍源代码,并给出具体的例子。第二篇会告诉你如何对一个 MergeSort 进行优化,介绍了 TimSort 背后的思想。

推荐阅读

  • TimSort in Java 7
  • 理解timsort

转载于:https://www.cnblogs.com/Iambda/p/3933457.html

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