187. Repeated DNA Sequences

题目：

All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: "ACGAATTCCG". When studying DNA, it is sometimes useful to identify repeated sequences within the DNA.

Write a function to find all the 10-letter-long sequences (substrings) that occur more than once in a DNA molecule.

For example,

Given s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT",

Return:
["AAAAACCCCC", "CCCCCAAAAA"].

链接： http://leetcode.com/problems/repeated-dna-sequences/　　

题解：

求repeating molecule of DNA sequence。最直接的想法就是从头遍历，把substring加入到HashMap中，然后进行比较。这样的话因为substring()的复杂度是O(n)，所以整个算法复杂度是O(n²)。看到有讨论用Rabin-Karp的Rolling Hash自己做Hash Function。这样做的好处我觉得可能是减少了substring()的次数，但总得来说时间复杂度也还是O(n²)。而且做了Rabin-Karp的话要不要要不要判断collision，collision以后用Monte-carlo还是Las Vegas检测结果，也是问题。要再研究一下。

Time Complexity - O(n²)， Space Complexity - O(n)。

public class Solution {
    public List<String> findRepeatedDnaSequences(String s) {
        List<String> res = new ArrayList<String>();
        if(s == null || s.length() < 10)
            return res;
        Map<String, Integer> map = new HashMap<>();

        for(int i = 0; i < s.length() - 9; i++) {
            String subStr = s.substring(i, i + 10);
            if(map.containsKey(subStr)) {
                if(map.get(subStr) == 1)
                    res.add(subStr);
                map.put(subStr, map.get(subStr) + 1);
            } else
                map.put(subStr, 1);
        }

        return res;
    }
}

二刷:

还是用的老方法，利用HashMap进行比较。

看到Stefan的用Set也能做，更像Python的风格，速度更快。

Rolling Hash的话，Time Complexity是O(n)，但每次都要计算10个数的hash value，速度也不快。有机会的话联系一下好了。

Java:

HashMap:

Time Complexity - O(n²)， Space Complexity - O(n)。

public class Solution {
    public List<String> findRepeatedDnaSequences(String s) {
        List<String> res = new ArrayList<>();
        if (s == null) return res;
        Map<String, Integer> map = new HashMap<>();
        for (int i = 0; i + 10 <= s.length(); i++) {
            String str = s.substring(i, i + 10);
            if (!map.containsKey(str)) {
                map.put(str, 1);
            } else {
                if (map.get(str) == 1) res.add(str);
                map.put(str, 2);
            }
        }
        return res;
    }
}

Reference：

https://leetcode.com/discuss/24595/short-java-rolling-hash-solution

https://leetcode.com/discuss/24557/just-7-lines-of-code

https://leetcode.com/discuss/24478/i-did-it-in-10-lines-of-c

https://leetcode.com/discuss/25399/clean-java-solution-hashmap-bits-manipulation

https://leetcode.com/discuss/25536/am-understanding-the-problem-wrongly-what-about-aaaaccccca

https://leetcode.com/discuss/29623/11ms-solution-with-unified-hash-fxn

https://leetcode.com/discuss/54777/easy-to-understand-java-solution-with-well-commented-code

https://leetcode.com/discuss/46948/accepted-java-easy-to-understand-solution

https://leetcode.com/discuss/64841/7-lines-simple-java-o-n