最近看了一些排序相关的文章，因此比较好奇，Java中的排序是如何做的。本片文章介绍的是JDK1.8，List中的sort方法。
先来看看List中的sort是怎么写的：
@SuppressWarnings({"unchecked", "rawtypes"}) 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); } }
首先，你需要传入一个比较器作为参数，这个好理解，毕竟你肯定要定一个比较标准。然后就是将list转换成一个数组，再对这个数组进行排序，排序完之后，再利用iterator重新改变list。
接着，我们再来看看Arrays.sort：
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); } } public static void sort(Object[] a) { if (LegacyMergeSort.userRequested) legacyMergeSort(a); else ComparableTimSort.sort(a, 0, a.length, null, 0, 0); } static final class LegacyMergeSort { private static final boolean userRequested = java.security.AccessController.doPrivileged( new sun.security.action.GetBooleanAction( "java.util.Arrays.useLegacyMergeSort")).booleanValue(); }
这样可以看出，其实排序的核心就是TimSort，LegacyMergeSort大致意思是表明如果版本很旧的话，就用这个，新版本是不会采用这种排序方式的。
我们再来看看TimSort的实现：
private static final int MIN_MERGE = 32; 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) { // 获得最长的递增序列 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; }
如果小于2个，代表不再不需要排序；如果小于32个，则采用优化的二分排序。怎么优化的呢？首先获得最长的递增序列：
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 // 一开始是递减序列，就找出最长递减序列的最后一个下标 while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0) runHi ; // 逆转前面的递减序列 reverseRange(a, lo, runHi); } else { // Ascending while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0) runHi ; } return runHi - lo; }
接着进行二分排序：
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). */ // start位置是递增序列后的第一个数的位置 // 从前面的递增序列中找出start位置的数应该处于的位置 while (left < right) { // >>> 无符号右移 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 // 比pivot大的数往后移动一位 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; } }
好了，待排序数量小于32个的讲完了，现在来说说大于等于32个情况。首先，获得一个叫minRun的东西，这是个啥含义呢：
int minRun = minRunLength(nRemaining); 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) { // 这里我没搞懂的是为什么不直接将(n & 1)赋值给r，而要做一次逻辑或。 r |= (n & 1); n >>= 1; } return n r; }
各种位运算符，MINMERGE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，直到小于MINMERGE，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。其实看注释比较容易理解：
Returns the minimum acceptable run length for an array of the specified length. Natural runs shorter than this will be extended with 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.
返回结果其实就是用于接下来的合并排序中。
接下来就是一个while循环
do { // Identify next run // 获得一个最长递增序列 int runLen = countRunAndMakeAscending(a, lo, hi, c); // If run is short, extend to min(minRun, nRemaining) // 如果最长递增序列 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 // lo——runLen为将要被归并的范围 ts.pushRun(lo, runLen); // 归并 ts.mergeCollapse(); // Advance to find next run lo = runLen; nRemaining -= runLen; } while (nRemaining != 0);
这样，假设你的每次归并排序的两个序列为r1和r2，r1肯定是有序的，r2也已经被排成递增序列了，因此这样的归并排序就比较特殊了。
为什么要用归并排序呢，因为归并排序的时间复杂度永远为O(nlogn)，空间复杂度为O(n)，以空间换取时间。
好了，以上就是针对Java中的排序做的一次总结，但具体的归并代码还没有分析，其实我自己也没有完全研究透，为什么minRun的取值是这样的，这也和TimSort中的stackLen有关，有兴趣的小伙伴可以在下方留言，我们可以一起探讨。
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