[LeetCode] 187. Repeated DNA Sequences Java

题目:

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"].

 题意及分析：给出一个字符串，其中只有A,C,G,T四个字母，每10个字母作为一个子字符串，要求找到出现不止一次的子字符串。这道题直接用hastable方法求解，遍历字符串，对每个子字符串做判断，若在hashtable中不存在，就添加进去；若存在，如果出现的次数为1，那么将其添加进结果中，并更新出现次数，否则继续遍历。还有一种方法是将a,c,g,t使用3位bit来保存，然后10个字母，就30bit，这样就可以用一个整数来保存。

代码：

import java.util.ArrayList;
import java.util.Hashtable;
import java.util.List;
public class Solution {
    public List<String> findRepeatedDnaSequences(String s) {
        List<String> res = new ArrayList<>();
        Hashtable<String,Integer> temp = new Hashtable<>();

        for(int i=0;i<s.length()-9;i++){      //将每一个长度为10的子字符串进行遍历，没有就将其放进hashtable里面，有且现在之出现了一次就添加进结果里面。
            String subString = s.substring(i,i+10);
            if(temp.containsKey(subString)){
                int count=temp.get(subString);  //如果为1，则添加进结果，否则继续遍历
                if(count==1){
                    temp.remove(subString);
                    temp.put(subString,2);
                    res.add(subString);
                }
            }else{
                temp.put(subString,1);
            }
        }
        return res;
    }
}