Arrays.sort和Collections.sort实现原理解析

Arrays.sort和Collections.sort实现原理解析

1、使用

  • 排序

2、原理

  1. 事实上Collections.sort方法底层就是调用的array.sort方法,而且不论是Collections.sort或者是Arrays.sort方法,

  2. 跟踪下源代码吧,首先我们写个demo

public static void main(String[] args) {
List<String> strings = Arrays.asList("6", "1", "3", "1","2"); Collections.sort(strings);//sort方法在这里 for (String string : strings) { System.out.println(string);
}
}

简单得不能再简单的方法了,让我们一步步跟踪

  1. OK,往下面看,发现collections.sort方法调用的list.sort

Arrays.sort和Collections.sort实现原理解析

  1. 然后跟踪一下,list里面有个sort方法,但是list是一个接口,肯定是调用子类里面的实现,这里我们demo使用的是一个Arrays.asList方法,所以事实上我们的子类就是arraylist了。OK,看arraylist里面sort实现,选择第一个,为什么不选择第二个呢?(可以看二楼评论,解答得很正确,简单说就是用Arrays.sort创建的ArrayList对象)

Arrays.sort和Collections.sort实现原理解析

  1. OK,发现里面调用的Arrays.sort(a, c); a是list,c是一个比较器,我们来看一下这个方法 
    Arrays.sort和Collections.sort实现原理解析

我们没有写比较器,所以用的第二项,LegacyMergeSort.userRequested这个bool值是什么呢? 
跟踪这个值,我们发现有这样的一段定义:

  > Old merge sort implementation can be selected (for
> compatibility with broken comparators) using a system property.
> Cannot be a static boolean in the enclosing class due to
> circular dependencies. To be removed in a future release. 反正是一种老的归并排序,不用管了现在默认是关的
  1. OK,我们走的是sort(a)这个方法,接着进入这个 
    Arrays.sort和Collections.sort实现原理解析
  2. 接着看我们重要的sort方法
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的大小为0或者1就不用排了 // 当数组大小小于MIN_MERGE(32)的时候,就用一个"mini-TimSort"的方法排序,jdk1.7新加
if (nRemaining < MIN_MERGE) {
//这个方法比较有意思,其实就是将我们最长的递减序列,找出来,然后倒过来
int initRunLen = countRunAndMakeAscending(a, lo, hi);
//长度小于32的时候,是使用binarySort的
binarySort(a, lo, hi, lo + initRunLen);
return;
}
//先扫描一次array,找到已经排好的序列,然后再用刚才的mini-TimSort,然后合并,这就是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;
}
  1. 回到5,我们可以看到当我们写了比较器的时候就调用了TimSort.sort方法,源码如下
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;
}

和上面的sort方法是一样的,其实也就是TimSort的源代码

3、总结

不论是Collections.sort方法或者是Arrays.sort方法,底层实现都是TimSort实现的,这是jdk1.7新增的,以前是归并排序。TimSort算法就是找到已经排好序数据的子序列,然后对剩余部分排序,然后合并起来

 转载:https://blog.csdn.net/u011410529/article/details/56668545
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