java sort排序源码分析（TimSort排序）
入口：
default void sort(Comparator c) {
Object[] a = this.toArray();
Arrays.sort(a, (Comparator) c);
ListIterator i = this.listIterator();
for (Object e : a) {
i.next();
i.set((E) e);
}
} java排序方法调用的Arrays.sort ，传入两个参数，数据数组和comparator对象
public static void sort(T[] a, Comparator c) {
if (c == null) {
sort(a);
} else {
if (LegacyMergeSort.userRequested)
legacyMergeSort(a, c);
else
TimSort.sort(a, 0, a.length, c, null, 0, 0);
}
} 在sort方法中，有两种排序算法，传统排序，和TimSort
LegacyMergeSort.userRequested是使用jdk5的传统排序方法。
TimSort是改进后的归并排序，对归并排序在已经反向排好序的输入时表现为O(n^2)的特点做了特别优化。对已经正向排好序的输入减少回溯。对两种情况（一会升序，一会降序）的输入处理比较好（摘自百度百科）。
这里主要讲解TimSort排序
TimSort.sort(a, 0, a.length, c, null, 0, 0);static void sort(T[] a, int lo, int hi, Comparator c,
T[] work, int workBase, int workLen) 这里传入很多参数：a：数据数组，lo数据第一个元素索引，hi最后一个元素索引，c比较器对象，work工作空间数组，workBase工作空间可用空间，workLen工作集合的大小。
static void sort(T[] a, int lo, int hi, Comparator c,
T[] work, int workBase, int workLen) {
//断言错误情况
assert c != null && a != null && lo >= 0 && lo <= hi && hi <= a.length;
//判断数组长度是否小于2 如果是只有0或1，这种数组通常已经被排序
int nRemaining = hi - lo;
if (nRemaining < 2)
return; // Arrays of size 0 and 1 are always sorted
//如果数组长度小于MIN_MERGE(32)则使用二分排序
// 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;
} 长度小于32时的二分排序
1.数组从头开始寻找顺序片段，直到不满足要求；如果倒序也一直查找直到不满足要求，然后反转。
private static int countRunAndMakeAscending(T[] a, int lo, int hi,
Comparator c) {
assert lo < hi;
int runHi = lo + 1;
if (runHi == hi)
return 1;
//寻找数组中有序队列
// Find end of run, and reverse range if descending
if (c.compare(a[runHi++], a[lo]) < 0) { // Descending
while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0)
runHi++;
reverseRange(a, lo, runHi);
} else { // Ascending
while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0)
runHi++;
}
//返回有序片段长度
return runHi - lo;
} 2.使用二分查找来排序
private static void binarySort(T[] a, int lo, int hi, int start,
Comparator 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;
//查找到所需插入位置索引
while (left < right) {
int mid = (left + right) >>> 1;
if (c.compare(pivot, a[mid]) < 0)
right = mid;
else
left = mid + 1;
}
assert left == right;
//进行插入(插入位置是1或2时优化)
int n = start - left; // The number of elements to move
// Switch is just an optimization for arraycopy in default case
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;
}
} 这个相当于未分片的TimSort
长度大于32位时TimSort排序
1.计算出最小分片长度
/**
* 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 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;
计算出minRun，当n>=32时除2，直到小于32，（如果n为2的N幂，计算出来为16，否则保留最后五位加最后一次移位的r）
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) {
//&1之后，n为奇数则为1，偶数为0
r |= (n & 1);
//右移，相当于除2
n >>= 1;
}
return n + r;
}2.do-while
2.1取得最小升序片段长度(如果是降序则反转)，这个方法前面写到过
// Identify next run
int runLen = countRunAndMakeAscending(a, lo, hi, c);2.2如果该长度小于最小分片长度，则用二分查找插入变成满足最小分片长度的升序片段
// 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;
}2.3将该序列的起始位置和长度入栈
private void pushRun(int runBase, int runLen) {
this.runBase[stackSize] = runBase;
this.runLen[stackSize] = runLen;
stackSize++;
}2.4合并以有有序片段
private void mergeCollapse() {
while (stackSize > 1) {
int n = stackSize - 2;
//第一个片段长度小于后两个相加
if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) {
//如果小于后面第二个长度
if (runLen[n - 1] < runLen[n + 1])
//则将合并位置减一
n--;
mergeAt(n);
} else if (runLen[n] <= runLen[n + 1]) {
mergeAt(n);
} else {
break; // Invariant is established
}
}
}合并操作，先查出来两个片段边界元素在另外片段的位置
private void mergeAt(int i) {
assert stackSize >= 2;
assert i >= 0;
assert i == stackSize - 2 || i == stackSize - 3;
//数据初始化
int base1 = runBase[i];
int len1 = runLen[i];
int base2 = runBase[i + 1];
int len2 = runLen[i + 1];
assert len1 > 0 && len2 > 0;
assert base1 + len1 == base2;
/*
* 记录合并后的序列的长度
*/
runLen[i] = len1 + len2;
if (i == stackSize - 3) {
runBase[i + 1] = runBase[i + 2];
runLen[i + 1] = runLen[i + 2];
}
stackSize--;
/*
* 查找到run2的第一个元素排序在run1的位置
*/
int k = gallopRight(a[base2], a, base1, len1, 0, c);
assert k >= 0;
base1 += k;
len1 -= k;
if (len1 == 0)
return;
/*
* 查找到run1最后一个元素排序在run2的位置
*/
len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c);
assert len2 >= 0;
if (len2 == 0)
return;
//合并操作
// Merge remaining runs, using tmp array with min(len1, len2) elements
if (len1 <= len2)
mergeLo(base1, len1, base2, len2);
else
mergeHi(base1, len1, base2, len2);
}找到两个位置之后，则只需归并中间的字段
合并方法代码
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 c = this.c; // Use local variable for performance
int minGallop = this.minGallop; // " " " " "
outer:
while (true) {
int count1 = 0; // Number of times in a row that first run won
int count2 = 0; // Number of times in a row that second run won
/*
* Do the straightforward thing until (if ever) one run starts
* winning consistently.
*/
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);
}
}这段代码合并代码步骤
2.4.1分配临时片段，用于合并
2.4.2计数count整段合并
注：这里当len1=0抛出异常：Comparison method violates its general contract!，这是在整段合并时，识别到run1有片段应该合并到run2起始位置；但是在合并之前有过判断run1中小于run2第一个元素的片段已经不在合并范围内了，那么合并的run1不可能有片段还在run2的起始值之前（可以看合并的图示更好理解）。所以大家在重写compare方法时需要考虑周全。
以上是对java中timsort排序的一些浅显的解读。