Arrays.sort排序算法

一时失言乱红尘 2024-04-18 19:52 5阅读 0赞

java默认的排序算法及思路  
List.sort() 还是 Collections.sort底层的排序都是调用了Arrays.sort()方法

Arrays.sort方法如下：

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);
        }
    }

主要提供了两种排序算法，mergeSort 和  TimSort.sort  
mergeSort思想：归并排序与合并+插入排序，存在问题：对已经反向排好序的输入时复杂度为O(n^2)  
timesort思想：先合理分区（两种：时间轴正序或反序），然后再对反序的分区处理，最后进行合并，合并先合并小的，但是增加了针对正向和反向和大规模数据的优化处理。解决上面mergesort的问题。

什么时候用mergesort，什么时候用那个timesort?  
LegacyMergeSort.userRequested的字面意思大概就是“用户请求传统归并排序”的意思，通过：  
System.setProperty("java.util.Arrays.useLegacyMergeSort", "true");  
现在已经默认采用timesort排序  
归并算法：归并排序与合并+插入排序（分到最后数据量较小采用）

private static void mergeSort(Object[] src,
                                  Object[] dest,
                                  int low,
                                  int high,
                                  int off) {
        int length = high - low;
    
        // Insertion sort on smallest arrays
        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;
        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.
        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
        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;
    }

timesort 充分运用已有有序序列，合理分区+反转部分分区+合并 ，思想如下：  
分区  
分区的思想是扫描一次数组，把连续正序列（如果是升序排序，那么正序列就是升序序列）（也就是后面所指的run），如果是反序列，把分区里的元素反转一下。 例如  
1，2，3，6，4，5，8，6，4 划分分区结果为  
\[1,2,3,6\],\[4,5,8\],\[6,4\]  
然后反转反序列  
\[1,2,3,6\],\[4,5,8\],\[4,6\]

合并  
考虑一个极端的例子，比如分区的长度分别为 10000，10，1000，10，10，我们当然希望是先让10个10合并成20， 20和1000合并成1020如此下去， 如果从从左往右顺序合并的话，每次都用到10000这个数组和去小的数组合并，代价太大了。所以我们可以用一个策略来优化合并的顺序。

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) {
            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;
    }

获取单次运行的方法如下：

/**
     * Returns the minimum acceptable run length for an array of the specified
     * length. Natural runs shorter than this will be extended with
     * {@link #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.
     *
     * @param n the length of the array to be sorted
     * @return the length of the minimum run to be merged
     */
    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) {
            r |= (n & 1);
            n >>= 1;
        }
        return n + r;
    }

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).
             */
            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;
        }
    }

保存处理结果与合并代码如下：

/**
     * Pushes the specified run onto the pending-run stack.
     *
     * @param runBase index of the first element in the run
     * @param runLen  the number of elements in the run
     */
    private void pushRun(int runBase, int runLen) {
        this.runBase[stackSize] = runBase;
        this.runLen[stackSize] = runLen;
        stackSize++;
    }
    
    /**
     * 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
            }
        }
    }