Collections.sort内部原理

先看源码

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

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

Collections.sort 共有两个方法，参数不同，实际调用了list.sort方法

继续点进去看

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);
    }
}

最终实现

可以看到，sort方法实际由 Arrays.sort实现。

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);
    }
}

不带Comparator的实现，sort方法

public static void sort(Object[] a) {
    if (LegacyMergeSort.userRequested)
        legacyMergeSort(a);
    else
        ComparableTimSort.sort(a, 0, a.length, null, 0, 0);
}

来看 legacyMergeSort(a)

private static <T> void legacyMergeSort(T[] a, Comparator<? super T> c) {
    T[] aux = a.clone();
    if (c==null)
        mergeSort(aux, a, 0, a.length, 0);
    else
        mergeSort(aux, a, 0, a.length, 0, c);
}

mergeSort

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++];
    }
}

这里是最终的方法

来看分支 TimSort.sort

TimSort.sort

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;
}