Java1.7之后Arrays.sort对数组排序DualPivotQuicksort.sort

Posted 2021-12-28 java李杨勇

有粉丝叫我帮他做一下这道题。。。。

 额。。。。。。这同学应该好好听课啦 哈哈

 int[] a = 25, 24, 12, 76, 101, 96, 28;
        Arrays.sort(a);//排序
        System.out.println("排序后数组如下");
        for (int i = 0; i < a.length; i++) 
            System.out.print(a[i] + " ,");
        

既然写到这里了就看下底层实现吧

断点跟踪调用的是DualPivotQuicksort.java类的java双基准快速排序方法sort实现

 跟踪进去就是具体排序方法的实现、其中具体方法：参数 int[] a是需被排序的int数组, left和right是该数组中需要被排序的部分的左右界限. 而后面的work, workBase和workLen三个参数其实并不会参与双基准快速排序, 而是当系统认为本数组更适合使用归并排序(merge sort)的时候, 供归并排序使用。具体代码算法源码如下。

static void sort(int[] a, int left, int right,
                     int[] work, int workBase, int workLen) 
        // Use Quicksort on small arrays
        if (right - left < QUICKSORT_THRESHOLD) 
            sort(a, left, right, true);
            return;
        

        /*
         * Index run[i] is the start of i-th run
         * (ascending or descending sequence).
         */
        int[] run = new int[MAX_RUN_COUNT + 1];
        int count = 0; run[0] = left;

        // Check if the array is nearly sorted
        for (int k = left; k < right; run[count] = k) 
            if (a[k] < a[k + 1])  // ascending
                while (++k <= right && a[k - 1] <= a[k]);
             else if (a[k] > a[k + 1])  // descending
                while (++k <= right && a[k - 1] >= a[k]);
                for (int lo = run[count] - 1, hi = k; ++lo < --hi; ) 
                    int t = a[lo]; a[lo] = a[hi]; a[hi] = t;
                
             else  // equal
                for (int m = MAX_RUN_LENGTH; ++k <= right && a[k - 1] == a[k]; ) 
                    if (--m == 0) 
                        sort(a, left, right, true);
                        return;
                    
                
            

            /*
             * The array is not highly structured,
             * use Quicksort instead of merge sort.
             */
            if (++count == MAX_RUN_COUNT) 
                sort(a, left, right, true);
                return;
            
        

        // Check special cases
        // Implementation note: variable "right" is increased by 1.
        if (run[count] == right++)  // The last run contains one element
            run[++count] = right;
         else if (count == 1)  // The array is already sorted
            return;
        

        // Determine alternation base for merge
        byte odd = 0;
        for (int n = 1; (n <<= 1) < count; odd ^= 1);

        // Use or create temporary array b for merging
        int[] b;                 // temp array; alternates with a
        int ao, bo;              // array offsets from 'left'
        int blen = right - left; // space needed for b
        if (work == null || workLen < blen || workBase + blen > work.length) 
            work = new int[blen];
            workBase = 0;
        
        if (odd == 0) 
            System.arraycopy(a, left, work, workBase, blen);
            b = a;
            bo = 0;
            a = work;
            ao = workBase - left;
         else 
            b = work;
            ao = 0;
            bo = workBase - left;
        

        // Merging
        for (int last; count > 1; count = last) 
            for (int k = (last = 0) + 2; k <= count; k += 2) 
                int hi = run[k], mi = run[k - 1];
                for (int i = run[k - 2], p = i, q = mi; i < hi; ++i) 
                    if (q >= hi || p < mi && a[p + ao] <= a[q + ao]) 
                        b[i + bo] = a[p++ + ao];
                     else 
                        b[i + bo] = a[q++ + ao];
                    
                
                run[++last] = hi;
            
            if ((count & 1) != 0) 
                for (int i = right, lo = run[count - 1]; --i >= lo;
                    b[i + bo] = a[i + ao]
                );
                run[++last] = right;
            
            int[] t = a; a = b; b = t;
            int o = ao; ao = bo; bo = o;
        
    

虽然看似简单的一行代码其实底层的实现也是很复杂的、对算法感兴趣的同学可以仔细看看底层实现。