leetcode 187. Repeated DNA Sequences 求重复的DNA串 ---------- java

2023-02-12,,,

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"].

其实就是一个字符串,然后以10个为单位,求重复两次以上的字符串。

1、用一个set就可以实现了。

public class Solution {
public List<String> findRepeatedDnaSequences(String s) {
List<String> list = new ArrayList();
int len = s.length();
if (len <= 10){
return list;
}
HashSet<String> set = new HashSet();
for (int i = 10; i <= len; i++){
String str = s.substring(i - 10, i);
if (set.contains(str) && !list.contains(str)){
list.add(str);
} else {
set.add(str);
}
}
return list;
}
}

2、discuss里面是有一些利用位操作的,例如。

public List<String> findRepeatedDnaSequences(String s) {
Set<Integer> words = new HashSet<>();
Set<Integer> doubleWords = new HashSet<>();
List<String> rv = new ArrayList<>();
char[] map = new char[26];
//map['A' - 'A'] = 0;
map['C' - 'A'] = 1;
map['G' - 'A'] = 2;
map['T' - 'A'] = 3; for(int i = 0; i < s.length() - 9; i++) {
int v = 0;
for(int j = i; j < i + 10; j++) {
v <<= 2;
v |= map[s.charAt(j) - 'A'];
}
if(!words.add(v) && doubleWords.add(v)) {
rv.add(s.substring(i, i + 10));
}
}
return rv;
}

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