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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"].
大致思路很简单，用一个hashmap来存储对应10个长度DNA的字符串及出现次数，最后将出现次数大于一次的存入list中，这里主要一个问题是map的key如果直接用字符串，会出现exceed time limit问题，必须将该DNA字符串hash成一个int型整数，A->00;C->01;G->10;T->11;这样一个10个字符长度的DNA序列映射成一个20位的2进制数，可将该2进制数作为key。代码如下：
public class Solution { //将字符转换成对应2位2进制数 public int toInt(char c) { if(c==‘A‘) return 0; if(c==‘C‘) return 1; if(c==‘G‘) return 2; else return 3; } //将hashcode转换成DNA序列 public String tostring(int n) { StringBuffer sb = new StringBuffer(); for(int i=0;i<10;i++) { char c = ‘T‘; int temp = n%4; n = n>>2; if(temp==0) c = ‘A‘; if(temp==1) c = ‘C‘; if(temp==2) c = ‘G‘; sb.insert(0,c); } return sb.toString(); } public List<String> findRepeatedDnaSequences(String s) { List<String> re = new ArrayList<String>(); Map<Integer,Integer> map = new HashMap<Integer,Integer>(); int size = s.length(); if(size<=10) return re; int tmp = 0; for(int i=0;i<10;i++) { tmp = tmp<<2; tmp = tmp|toInt(s.charAt(i)); } map.put(tmp,1); for(int j=10;j<size;j++) { tmp = ((tmp&0x3ffff)<<2)|toInt(s.charAt(j));//先讲最高2位置0在左移两位 if(map.containsKey(tmp)) { map.put(tmp,map.get(tmp)+1); } else { map.put(tmp,1); } } Set<Integer> keys = map.keySet(); for(Integer key:keys) { if(map.get(key)>1) re.add(tostring(key)); } return re; } }
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原文地址：http://www.cnblogs.com/mrpod2g/p/4299559.html
