以下实施是稳定的,因为它在标记为XXX的行使用了<=
而不是<
。这也使它更有效率。是否有理由在此行使用<
而不是<=
?
/**
class for In place MergeSort
**/
class MergeSortAlgorithm extends SortAlgorithm {
void sort(int a[], int lo0, int hi0) throws Exception {
int lo = lo0;
int hi = hi0;
pause(lo, hi);
if (lo >= hi) {
return;
}
int mid = (lo + hi) / 2;
/*
* Partition the list into two lists and sort them recursively
*/
sort(a, lo, mid);
sort(a, mid + 1, hi);
/*
* Merge the two sorted lists
*/
int end_lo = mid;
int start_hi = mid + 1;
while ((lo <= end_lo) && (start_hi <= hi)) {
pause(lo);
if (stopRequested) {
return;
}
if (a[lo] <= a[start_hi]) { // LINE XXX
lo++;
} else {
/*
* a[lo] >= a[start_hi]
* The next element comes from the second list,
* move the a[start_hi] element into the next
* position and shuffle all the other elements up.
*/
int T = a[start_hi];
for (int k = start_hi - 1; k >= lo; k--) {
a[k+1] = a[k];
pause(lo);
}
a[lo] = T;
lo++;
end_lo++;
start_hi++;
}
}
}
void sort(int a[]) throws Exception {
sort(a, 0, a.length-1);
}
}
答案 0 :(得分:7)
因为代码中的<=
确保不会交换相同值的元素(在排序数组的左半部分和右半部分)。
而且,它避免了无用的交流。
if (a[lo] <= a[start_hi]) {
/* The left value is smaller than or equal to the right one, leave them as is. */
/* Especially, if the values are same, they won't be exchanged. */
lo++;
} else {
/*
* If the value in right-half is greater than that in left-half,
* insert the right one into just before the left one, i.e., they're exchanged.
*/
...
}
假设两半中的同值元素(例如,'5')和上面的运算符是<
。
如上面的评论所示,右边的'5'将插入左'5'之前,换句话说,将交换相同值的元素。
这意味着排序不稳定。
而且,交换同值元素也是低效的。
我认为效率低下的原因来自算法本身。 您的合并阶段是使用插入排序实现的(如您所知,它是O(n ^ 2))。
在对大型数组进行排序时,可能需要重新实现。
答案 1 :(得分:2)