@zero1036
2017-03-14T17:11:45.000000Z
字数 6055
阅读 4240
Java-Base
数据结构与算法
TimSort原理:现实数据通常会有部分是已经排好序,TimSort正是利用这一点,将数组拆成多个部分已排序的分区,部分未排序分区重新排序,最后将多个分区合并并排序。
例如:array[] = [24,63,70,55,41,92,81,80]
,排序步骤如下:
1. 拆分分区:[24,63]
, [70,55]
, [41,92]
, [81,80]
2. 重排分区:[24,63]
, [55,70]
, [41,92]
, [80,81]
3. 合并分区:[24,63,55,70]
, [41,92,80,81]
4. 重排分区:[24,55,63,70]
, [41,80,81,92]
5. 合并分区:[24,55,63,70,41,80,81,92]
6. 重排:[24,41,55,63,70,80,81,92]
JDK1.8以后默认采用Timsort排序,实现如下:
List list = new ArrayList<Integer>();
...
Collections.sort(list, new Comparator<Integer>() {
public int compare(Integer o1, Integer o2) {
if (o2 > o1) {
return 1;
} else {
return -1;
}
}
});
Java对于Timsort的实现与上述原理有区别。Java版首先会根据数组长度,采用Binarysort(折半插入排序法)对长度小于32(MIN_MERGE)直接进行排序返回结果;However,对于长度大于等于32的数组,先分区,再对单个分区进行采用Binarysort排序,最后合并分区并排序。
要点1:分区方法
假定数据长度为n,
If n < MIN_MERGE(32),返回n
Else if n 是2的倍数,返回MIN_MERGE(32) / 2 = 16
Else 返回整数k,k取值范围MIN_MERGE(32) / 2 = 16 <= k <= MIN_MERGE(32)
例:
Array Length = 15 ; minRun = 15
Array Length = 50 ; minRun = 25
Array Length = 500 ; minRun = 32
要点2:分区排序方法:二分法插入排序:
1、 二分法插入排序Binarysort要求首先找出数组(此数组即分区)中从0位开始连续升序区块,及区块下一位元素pivot;
例:
[1,3,5,7,9,4,8]
的起始连续升序区块是[1,3,5,7,9]
,区块长度为5,即runLen;pivot是4
;
2、 通过二分法比较pivot与区分元素的升降序关系,计算pivot在区块中的位置;并插入到该位置,组成新的区块;
例:
4
在区块中[1,3,5,7,9]
,先比较5 > 4
,是降序;再比较3 < 4
,是升序,确认位置,通过native方法System.arraycopy()
插入到区块中;
3、 再计算新区块与新pivot的位置关系,直到完成排序。
4、 注意:以上升序的定义是Comparator.compare(右元素, 左元素) >= 0
,反之为降序;非数值上的升降序。
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; // 长度为0或1数组无需排序
// 数组长度小于32时,直接采用binarySort排序,无需合并
if (nRemaining < MIN_MERGE) {
int initRunLen = countRunAndMakeAscending(a, lo, hi, c);
binarySort(a, lo, hi, lo + initRunLen, c);
return;
}
TimSort<T> ts = new TimSort<>(a, c, work, workBase, workLen);
// 获取分区长度
int minRun = minRunLength(nRemaining);
do {
// 计算目标数组指定范围中,连续升序或连续降序的元素组run(最少3个元素),并返回run长度
int runLen = countRunAndMakeAscending(a, lo, hi, c);
// 若分区中连续升降序的元素组长度 等于 分区长度,则无需排序;反之binarySort重排
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;
}
If n < MIN_MERGE(32),返回n(数据长度)
Else if n 是2的倍数,返回MIN_MERGE(32)/2=16
Else 返回整数k,k取值范围MIN_MERGE/2(16) <= k <= MIN_MERGE(32),且such that n/k
is close to, but strictly less than, an exact power of 2.不知如何理解?
private static int minRunLength(int n) {
assert n >= 0;
int r = 0;
while (n >= MIN_MERGE) {
r |= (n & 1);
n >>= 1;
}
return n + r;
}
/**
* 折半插入排序
* 排序结果为升序,Comparator.compare(右, 左) >= 0
* 例如:假设目标数组a长度为10,数组头3位已经排好升序,所以
* lo = 0
* hi = 9 + 1
* start = 3 (0 1 2位已排序,从第3位开始计算)
* @param a 目标数组
* @param lo 指定排序范围首个元素位置
* @param hi 指定排序范围最后元素位置 + 1
* @param start 从start位置开始计算排序,即lo到start部分不排序
*/
private static <T> void binarySort(T[] a, int lo, int hi, int start,
Comparator<? super T> c) {}
详细参考:插入排序分析与Java实现
计算目标数组指定范围中,连续升序或连续降序的元素组run(最少3个元素),并返回run长度,若连续降序run,则重置其中元素为升序(即a[lo] <= a[lo + 1] <= a[lo + 2] <=...
);
注意:如果run为严格降序,即run中的前一元素大于后一元素(a[lo] > a[lo + 1] > a[lo + 2] > ...
),可以元素翻转。严格降序为>
,>=
不是,如果在 >=
的情况下进行翻转这个算法就不再是stable。
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;
// 计算连续升序或降序的最后一位元素位置(runHi),降序则通过`reverseRange()`翻转元素为升序
// 首先比较第0与第2位
if (c.compare(a[runHi++], a[lo]) < 0) { // 降序
// 再从第1与第2位的比较开始,依次比较n与n + 1位,直到比较为非降序
while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0)
runHi++;
// 翻转数组指定范围的元素,lo:位置,runHi:高位位置,即范围长度
reverseRange(a, lo, runHi);
} else { // 升序
// 以上同理
while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0)
runHi++;
}
return runHi - lo;
}
lo:起始位置,从0起始
hi:指定范围长度
private static void reverseRange(Object[] a, int lo, int hi) {
hi--;
while (lo < hi) {
Object t = a[lo];
a[lo++] = a[hi];
a[hi--] = t;
}
}
private void mergeLo(int base1, int len1, int base2, int len2) {
assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
// Copy first run into temp array
T[] a = this.a; // For performance
T[] tmp = ensureCapacity(len1);
int cursor1 = tmpBase; // Indexes into tmp array
int cursor2 = base2; // Indexes int a
int dest = base1; // Indexes int a
System.arraycopy(a, base1, tmp, cursor1, len1);
// Move first element of second run and deal with degenerate cases
a[dest++] = a[cursor2++];
if (--len2 == 0) {
System.arraycopy(tmp, cursor1, a, dest, len1);
return;
}
if (len1 == 1) {
System.arraycopy(a, cursor2, a, dest, len2);
a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
return;
}
Comparator<? super T> c = this.c; // Use local variable for performance
int minGallop = this.minGallop; // " " " " "
outer:
while (true) {
int count1 = 0; // run1操作次数
int count2 = 0; // run2操作次数
/*
* 【逐一比对】
* run1(tmp)与run2(a第2段)根据升序逐一比较,当compare()<0时,写入到目标数组a
* 当连续采用run1或run2的元素次数达到(等于)7次(minGallop),则采用方法
* 例如:run1 = [1,6,7...] ; run2 = [2,4,8...]
* 结果:
* a = [1]
* a = [1,2]
* a = [1,2,4]
* a = [1,2,4,6]
* ...类推
*/
do {
assert len1 > 1 && len2 > 0;
if (c.compare(a[cursor2], tmp[cursor1]) < 0) {
a[dest++] = a[cursor2++];
count2++;
count1 = 0;
if (--len2 == 0)
break outer;
} else {
a[dest++] = tmp[cursor1++];
count1++;
count2 = 0;
if (--len1 == 1)
break outer;
}
} while ((count1 | count2) < minGallop);
/*
* One run is winning so consistently that galloping may be a
* huge win. So try that, and continue galloping until (if ever)
* neither run appears to be winning consistently anymore.
*/
do {
assert len1 > 1 && len2 > 0;
count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c);
if (count1 != 0) {
System.arraycopy(tmp, cursor1, a, dest, count1);
dest += count1;
cursor1 += count1;
len1 -= count1;
if (len1 <= 1) // len1 == 1 || len1 == 0
break outer;
}
a[dest++] = a[cursor2++];
if (--len2 == 0)
break outer;
count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c);
if (count2 != 0) {
System.arraycopy(a, cursor2, a, dest, count2);
dest += count2;
cursor2 += count2;
len2 -= count2;
if (len2 == 0)
break outer;
}
a[dest++] = tmp[cursor1++];
if (--len1 == 1)
break outer;
minGallop--;
} while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
if (minGallop < 0)
minGallop = 0;
minGallop += 2; // Penalize for leaving gallop mode
} // End of "outer" loop
this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field
if (len1 == 1) {
assert len2 > 0;
System.arraycopy(a, cursor2, a, dest, len2);
a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
} else if (len1 == 0) {
throw new IllegalArgumentException(
"Comparison method violates its general contract!");
} else {
assert len2 == 0;
assert len1 > 1;
System.arraycopy(tmp, cursor1, a, dest, len1);
}
}