Difference between revisions of "Talk:TADM2E 4.4"
From Algorithm Wiki
m |
|||
| Line 5: | Line 5: | ||
Each pass is O(n) since we have to visit each element, so the entire algorithm is O(n). | Each pass is O(n) since we have to visit each element, so the entire algorithm is O(n). | ||
| + | |||
| Line 17: | Line 18: | ||
if a[index].color=="blue" | if a[index].color=="blue" | ||
swap(a,++index_blue,index) | swap(a,++index_blue,index) | ||
| + | |||
| + | The ''swap''' above makes it a non-stable sort. | ||
Revision as of 07:31, 30 December 2016
Why break up the pairs and use buckets? It seems like you simply need three passes: First pass - get the reds and insert them in order into a new array of size n (assuming we can use additional space) Second pass - get the blues and insert them in order into the array Third pass - get the yellows and insert them in order into the array
Each pass is O(n) since we have to visit each element, so the entire algorithm is O(n).
EDIT: Why three passes ? .Two passes can do it
index_red=-1
for index =0 to a.length-1
if a[index].color == "Red"
swap(a,++index_red,index)
index_blue=index_red
for index = index_blue+1 to a.length-1
if a[index].color=="blue"
swap(a,++index_blue,index)
The swap' above makes it a non-stable sort.
