LeetCode5 Solved #49
LeetCode5 Solved #49
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| if(i+3 <= s.length() && isPalindrome(s.substring(i,i+3))) { | ||
| palindromeList.add(new Position_LeetCode5(i, i+3)); | ||
| } | ||
| } |
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I got what you were trying to do here.
You put all segments with length 2 or length 3 into the list, and then you can enumerate all palindromes with length 2+2k and 3+2k. In this case you can enumerate all lengths except for 1, but we can ignore 1 for sure.
However, I don't think your algorithm runs in O(N).
Imagine a case with a string input "aaaa........................................aa", you basically enumerate all the O(N^2) of possibilities, where N equals to the length of the string.
Please try:
- Carefully think about why your algorithm is O(N^2) instead of O(N).
- Try to create some test cases which make your code run slowly. (this is very important if you have an interview in big techs like Google) you can watch this
- In order to solve this problem in O(N), you'll need an algorithm called "Manacher's Algorithm". pseudocode in wikipedia
Even more:
Manacher's algorithm is an extended version of Z-algorithm. You can also try to learn the Z-algorithm. (I've never implemented this before, but it might be useful someday)
| int strLen = str.length(); | ||
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| // 각 위치에서 회문을 만들어주는 가장 큰 반지름을 저장하기 위한 배열 생성 | ||
| int strArr[] = new int[strLen]; |
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I'll recommend you to give this array a more meaningful name.
The name should clearly indicate that the value means the radius of a given position.
| // 이미 입력된 반지름이 있으면 거기부터 시작 | ||
| int radius = strArr[center]; | ||
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| while(center >= radius + 1 && // 왼쪽으로 중심에서 반지름까지가 0보다 크거나 같을 때 |
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It took me a while to understand what center >= radius + 1 means.
I'll recommend you to calculate the "position" in this case.
The center here means the position of the center, so the position of the left side should be center - radius.
Hence, the next coordinate you're trying should be center - radius - 1.
I think writing center - radius - 1 >= 0 gives us a better semantic to what you're going for here because center - radius - 1 and 0 are both coordinates.
| newCenter++; | ||
| } | ||
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| center++; // 다음 글자에서 다시 측정 ( 뒤에 추가로 오는 글자를 포함하여 또 회문일수 있으므로 재측정 ) |
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Line 41 to line 59 is wrong. Please check the algorithm again.
You can try some examples on paper and see why this part is wrong.
Time complexity : O(N)