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*187. Repeated DNA Sequences (hashmap, one for loop)(difference between subsequence & substring)

sequence value n-2 return hashset cga AS repeated des

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"]

Solution: count the frequency of 10 letter words

class Solution {
    //find all the 10-letter-long sequences that occur more than once in a DNA molecule
    public List<String> findRepeatedDnaSequences(String s) {
        
//substring -- subset n +n-1+...+1: n-k+1 List<String> res = new ArrayList<String>(); Map<String,Integer> map = new HashMap<String,Integer>(); int n = s.length(); int k =10; if(n < k) return res; for(int i = 0; i<=n-k; i++){//11-10 1
String sub = s.substring(i, i+k); if(map.containsKey(sub)){ map.put(sub, map.get(sub)+1); }else { map.put(sub, 1); } } for(Map.Entry<String, Integer> entry : map.entrySet()){ if(entry.getValue() >1){ res.add(entry.getKey()); } } return res; } }

Solution 2: two HashSet with a non-duplicate feature.

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))//if add then first time, else add it
            repeated.add(ten);
    }
    return new ArrayList(repeated);
}

subsequence & substring

subsequence: subset 2^n

substring: continous string : n+n-1+n-2+...+1

*187. Repeated DNA Sequences (hashmap, one for loop)(difference between subsequence & substring)