And you see- the print- "Couldn't find" executes everytime the code comes out of the loop. So as said by Siddhesh, you would have to use a flag (or even the same index- checking if its reached the end of the array) to print the message- "Couldn't Find"
Also, if this is real production code (and not just an exercise) - there are more efficient ways of sorting an array than the bubble sort algorithm that you're using. There are methods for sorting arrays in class java.util.Arrays, which implement an optimized quicksort algorithm.
Campbell Ritchie wrote:Only for primitives, I see from the API. For Object sorting, it uses a merge sort, because merge sort is "stable".
Aha, ok. "Stable" means that two elements that are equal will not switch order (they will be left in the original order that they were in in the array). Note that merge sort is still more efficient than bubble sort: the API docs mention that it's guaranteed O(N * log(N)) performance, whereas bubble sort has average and worst case performance of O(N^2).
Joined: Oct 13, 2005
I had always thought that sort() used merge sort throughout, but that is because I didn't read the whole of the API. That is the correct interpretation of "stable"; since primitives only have one value, there is no need for a stable sort. And bubble sort is only quicker for the very smallest arrays, otherwise it is notorious for its inefficiency.