Middle-题目96:187. Repeated DNA Sequences
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题目原文:
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”].
题目大意:
给出一个DNA序列(由’A’,’C’,’G’,’T’组成的字符串),求其中所有的在该DNA序列中出现至少两次的10位子序列。
题目分析:
方法一:(朴素解法,基于哈希表)构建一个HashMap,其中Key是原序列的所有的10位子序列(空间复杂度O(n)),value是这个序列的出现次数。如果发现某个key出现次数为2则加入列表(超过2则跳过,否则列表会出现重复串)
方法二:(基于Bit manipulation,来自discuss中大神的算法)分别把’A’,’C’,’G’,’T’编码为00,01,10,11.先计算前9个字符的编码,然后从第10个字符开始,去掉最高位的编码(与0xfffff求与),再加上最后一位的编码,放入集合中,如果集合重复了,则加入最终返回的列表。
源码:(language:java)
方法一:
public class Solution { public List<String> findRepeatedDnaSequences(String s) { List<String> list = new ArrayList<String>(); HashMap<String, Integer> dict = new HashMap<String, Integer>(); for(int i = 0;i<=s.length()-10;i++) { String substr = s.substring(i,i+10); if(!dict.containsKey(substr)) { dict.put(substr, 1); } else { int value = dict.get(substr) + 1; if(value>2) continue; dict.put(substr, value); if(value == 2) list.add(substr); } } return list; }}
方法二:
public class Solution { public List<String> findRepeatedDnaSequences(String DNA) { ArrayList<String> res = new ArrayList<String>(); if(DNA.length()<10) return res; HashSet<Integer> once = new HashSet<Integer>(); HashSet<Integer> twice = new HashSet<Integer>(); int[] map = new int[26]; map['A'-'A'] = 0; map['C'-'A'] = 1; map['G'-'A'] = 2; map['T'-'A'] = 3; int enc = 0; for(int i=0; i<9; ++i){ enc <<=2; enc |= map[DNA.charAt(i)-'A']; } for(int j=9; j<DNA.length(); ++j){ enc <<=2; enc &= 0xfffff; enc |= map[DNA.charAt(j)-'A']; if(!once.add(enc) && twice.add(enc)) res.add(DNA.substring(j-9,j+1)); } return res; }}
成绩:
方法一:40ms,beats 67.09%,众数38ms,5.66%
方法二:32ms,beats 95.90%
cmershen的碎碎念:
第二种方法有两个巧妙的处理方法是值得学习的:一是enc &= 0xfffff;表示去掉原字符串最高位编码,二是if(!once.add(enc) && twice.add(enc))代表这个编码第二次出现(虽然这个我还不太理解)
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