LeetCode 187. Repeated DNA Sequences

题目描述

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.

Example:

Input: s = “AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT”

Output: [“AAAAACCCCC”, “CCCCCAAAAA”]

AC代码1

public class _187FindRepeatedDnaSequences {
    public List<String> findRepeatedDnaSequences(String s) {
        List<String> resList = new ArrayList<>();
        Set<Integer> words = new HashSet<>();
        Set<Integer> doublewords = new HashSet<>(); // 防止添加到resList中的字符串重复
        char[] maps = new char[26];
        //maps['A' - 'A'] = 0; 默认操作,无需显式
        maps['C' - 'A'] = 1;
        maps['G' - 'A'] = 2;
        maps['T' - 'A'] = 3;
        for(int i = 0;i + 9 < s.length();i++){ // 注意是 i + 9
            int val = 0;
            for(int j = 0;j < 10;j++){
                val <<= 2;
                val |= maps[s.charAt(j + i) - 'A'];
            }
            if(!words.add(val) && doublewords.add(val))
                resList.add(s.substring(i,i + 10));
        }
        return resList;
    }
}

AC代码2

public List<String> findRepeatedDnaSequences(String s) {
    Set seen = new HashSet(), repeated = new HashSet();
    for (int i = 0; i + 9 < s.length(); i++) {
        String ten = s.substring(i, i + 10);
        if (!seen.add(ten))
            repeated.add(ten);
    }
    return new ArrayList(repeated);
}

注:题目及解法均来自于LeetCode

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转载自blog.csdn.net/tkzc_csk/article/details/81981455