来自于:
http://svn.python.org/projects/python/trunk/Objects/listsort.txt
和:
http://en.wikipedia.org/wiki/Timsort
我看到Timsort在a0>a1>a2>...
的情况下有一些优化,但是对于下一个数组呢:
10000,10000,9999,9999,9998,9998,....,9,9,8,8,7,7,6,6,5,5,4,4,3,3,2,2,1,1,0,0
这种数组的时间效率是多少?
(为了简单起见,使用整数。需要稳定排序)我已经做过一些测量,并且似乎这样的数组不是Timsort的“好”情况。
实际上,JDK中的TimSorthttp://cr.openjdk.java.net/~martin/webrevs/openjdk7/timsort/raw_files/new/src/share/classes/java/util/TimSort.java 有一个方法“countRunAndMakeAscending”
@SuppressWarnings("unchecked")
private static int countRunAndMakeAscending(Object[] a, int lo, int hi) {
assert lo < hi;
int runHi = lo + 1;
if (runHi == hi)
return 1;
// Find end of run, and reverse range if descending
if (((Comparable) a[runHi++]).compareTo(a[lo]) < 0) { // Descending
while(runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) < 0)
runHi++;
reverseRange(a, lo, runHi);
} else { // Ascending
while (runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) >= 0)
runHi++;
}
return runHi - lo;
}
为什么不用另一种方式来实现它:
private static int countRunAndMakeAscending(Object[] a, int lo, int hi) {
int runHi = lo;
int lastEqual = lo;
int ascending = 0;
while (++runHi < hi) {
int c = ((Comparable) a[runHi+1]).compareTo(a[runHi]);
if (ascending == 0) {
if (c != 0) {
if (c > 0) {
ascending = 1;
} else {
ascending = -1;
reverseRange(a, lastEqual, runHi);
lastEqual = runHi;
}
}
} else if (ascending == 1) {
if (c < 0) {
return runHi - lo;
}
} else {
if (c > 0) {
reverseRange(a, lastEqual, runHi);
reverseRange(a, lo, runHi);
return runHi - lo;
} else if (c < 0) {
reverseRange(a, lastEqual, runHi);
lastEqual = runHi;
}
}
}
if (ascending == -1) {
reverseRange(a, lastEqual, runHi);
reverseRange(a, lo, runHi);
}
return runHi - lo;
}
那么它可以在非升序的情况下正常工作吗?
[(1,2),(2,0),(1,3)]
,如果排序结果是[(1,3),(1,2),(2,0)]
,那将是不好的。 - Oscar Smith