leetcode_Repeated DNA Sequences

描述:

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"].

思路:

1.很显然,暴力求解也是一种方法,尽管该方法是不可能的。

2.我们首先来看字母 ”A" "C" “G" "T" 的ASCII码,分别是65, 67, 71, 84,二进制表示为 1000001, 1000011, 1000111, 1010100。可以看到它们的后三位是不同,所以用后三位就可以区分这四个字母。一个字母用3bit来区分,那么10个字母用30bit就够了。用int的第29~0位分表表示这0~9个字符,然后把30bit转化为int作为这个子串的key,放入到HashTable中,以判断该子串是否出现过。

代码:

 public List<String> findRepeatedDnaSequences(String s)
	{
		List<String>list=new ArrayList<String>();
		int strLen=s.length();
		if(strLen<=10)
			return list;
		HashMap<Integer, Integer>map=new HashMap<Integer,Integer>();
		int key=0;
		for(int i=0;i<strLen;i++)
		{
			key=((key<<3)|(s.charAt(i)&0x7))&0x3fffffff;//k<<3,key左移3位,也就是将最左边的字符移除
			//s.charAt(i)&0x7)获得用于标记s.charAt(i)字符的低3位
			//&0x3fffffff抹去key左移三位后多出的高位不相关比特位
			if(i<9)continue;
			if(map.get(key)==null)//如果没有该整数表示的字符串,将其添加进map中
				map.put(key, 1);
			else if(map.get(key)==1)//如果存在,说明存在重复字符串并将其添加进结果list中
			{
				list.add(s.substring(i-9,i+1));
				map.put(key, 2);//防止重复添加相同的字符串
			}
		}
		return list;
	}

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时间: 2024-10-12 21:50:45

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