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Hello guys, i have a solution for this task and it works, but i think its not very elegant. I add at every second, so uneven, position the word "ham". If the last position is even, i manually add "ham" again. It works but as i said i dont like it very much. How were your solutions? As my code is working i cannot post it here. Thanks already for discussion

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Thomas

Level 41 , Bayreuth, Germany

11 Mai 2020, 14:14

I have two solutions 1st.. not very elegant:

for (int i = 0; i < list.size(); i++) {
    list.add(i*2 + 1, "Ham");
}

this one I like much more

for (int i = list.size(); i > 0; i--) {
    list.add(i, "Ham");
}

niobe92

Level 12 , Dresden, Germany

11 Mai 2020, 15:14

ah yeah the second one looks good. But shouldnt i be i = i-2; But anyway, thats the solution i was searching for. Thanks :)

Thomas

Level 41 , Bayreuth, Germany

11 Mai 2020, 17:07

Just try it and you'll see that it's i-- ;) Just think about it. If you'd just add elements it'll be like a loop -1. With the additional i-- you match every second entry.

niobe92

Level 12 , Dresden, Germany

13 Mai 2020, 09:30

Yes, youre right. The already existing word at the position goes to the next one and it fits. Perfect. THanks!

Ashish RajAnand

Level 13 , Bhilai , India

11 Mai 2020, 13:31

same problem bro we can't post solution. another method is just go help section and search resolve question or many comments question . You must look what mistake there and how different they coded from your.