Repeated DNA Sequences

原文地址：http://blog.csdn.net/u013325815/article/details/43601367

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"].

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Have you met this question in a real interview?

由于直接将字符串存入字典会导致Memory Limit Exceeded，采用位操作将字符串转化为整数可以减少内存开销

字典+位运算

//A=0x41, C=0x43, G=0x47, T=0x54

//A=0101, C=0103, G=0107, T=0124

//A,C,G,T最后3bit不同，即与mask“0111”作与运算可唯一标识ACGT。

[java] view plaincopy

1. public class Solution {  
2.     public List<String> findRepeatedDnaSequences(String s) {  
3.         List<String> list = new ArrayList<String>();  
4.         if(s == null || s.length()<=10) return list;  
5.           
6.         int mask = 0x7FFFFFF;  
7.         int i=0;  
8.         int cur = 0;  
9.         HashMap<Integer,Integer> hashmap = new HashMap<Integer,Integer>();  
10.         while(i<9){  
11.             cur = ((cur<<3) | s.charAt(i) & 7);  
12.             i++;  
13.         }  
14.           
15.         while(i<s.length()){  
16.             cur = ((cur & mask)<<3 | s.charAt(i) & 7);  
17.             i++;  
18.             if(hashmap.containsKey(cur)){  
19.                 int count = hashmap.get(cur);  
20.                 if(count == 1){  
21.                     list.add(s.substring(i-10,i));  
22.                 }  
23.                 hashmap.put(cur, count+1);  
24.             }else{  
25.                 hashmap.put(cur,1);  
26.             }  
27.         }  
28.         return list;  
29.     }  
30. }