public static void main(String[] args) {
         List<String> strings = Arrays.asList("6", "1", "3", "1","2");

         Collections.sort(strings);//sort方法在這里

         for (String string : strings) {

           System.out.println(string);
         }
       }

static void sort(Object[] a, int lo, int hi, Object[] work, int workBase, int workLen) {
         assert a != null && lo >= 0 && lo <= hi && hi <= a.length;

         int nRemaining = hi - lo;
         if (nRemaining < 2)
           return; // array的大小為0或者1就不用排了

         // 當(dāng)數(shù)組大小小于MIN_MERGE(32)的時候，就用一個"mini-TimSort"的方法排序，jdk1.7新加
         if (nRemaining < MIN_MERGE) {
          //這個方法比較有意思，其實就是將我們最長的遞減序列，找出來，然后倒過來
           int initRunLen = countRunAndMakeAscending(a, lo, hi);
           //長度小于32的時候，是使用binarySort的
           binarySort(a, lo, hi, lo + initRunLen);
           return;
         }
        //先掃描一次array，找到已經(jīng)排好的序列，然后再用剛才的mini-TimSort，然后合并，這就是TimSort的核心思想
         ComparableTimSort ts = new ComparableTimSort(a, work, workBase, workLen);
         int minRun = minRunLength(nRemaining);
         do {
           // Identify next run
           int runLen = countRunAndMakeAscending(a, lo, hi);

           // 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);
             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;
     }

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;
   }