public class Arrays{public static <T> void sort(T[] a, Comparator<? super T> c) {if (c == null) {sort(a);} else {if (LegacyMergeSort.userRequested)legacyMergeSort(a, c);elseTimSort.sort(a, 0, a.length, c, null, 0, 0);}}
}

public class Test1 {public static void main(String[] args) {Integer []data ={12,2,3,2,4,5,1};
// 实现降序排序，返回-1放左边，1放右边，0保持不变Arrays.sort(data, (str1, str2) -> {if (str1.compareTo(str2) > 0) {return -1;} else {return 1;}});System.out.println(Arrays.toString(data)); //[12, 5, 4, 3, 2, 2, 1]
    }
}

class TimSort<T> {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 mergesif (nRemaining < MIN_MERGE) {int initRunLen = countRunAndMakeAscending(a, lo, hi, c);binarySort(a, lo, hi, lo + initRunLen, c);return;}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 descendingif (c.compare(a[runHi++], a[lo]) < 0) { // Descendingwhile (runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0)runHi++;reverseRange(a, lo, runHi);} else {                              // Ascendingwhile (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0)runHi++;}return runHi - lo;}
}