题解 | #重复的DNA序列#

import java.util.*;


public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param DNA string字符串 1
     * @return string字符串一维数组
     */
    public String[] repeatedDNA (String DNA) {
        // write code here
        HashMap<String, Integer> hashMap = new HashMap<>();
        ArrayList<String> arrayList = new ArrayList<>();
        int len = DNA.length();
        for (int i = 0; i <= len - 10; i++) {
            String tmpStr = DNA.substring(i, i + 10);
            if (!isValid(tmpStr)) {
                continue;
            }
            else {
                int num = hashMap.getOrDefault(tmpStr, 0);
                num++;
                hashMap.put(tmpStr, num);
            }
        }
        // 此时，HashMap 存放了所有的合法子串出现的次数
        // 我们再重新遍历一次，将出现 2 次的子串添加到结果集中去
        for (int i = 0; i <= len - 10; i++) {
            String tmpStr = DNA.substring(i, i + 10);
            int num = hashMap.getOrDefault(tmpStr, 0);
            if (num > 1 && (!arrayList.contains(tmpStr))) {
                arrayList.add(tmpStr);
            }
        }
        String[] res = new String[arrayList.size()];
        for (int i = 0; i < arrayList.size(); i++) {
            res[i] = arrayList.get(i);
        }
        return res;
    }
    
    public boolean isValid(String str) {
        boolean bool = true;
        for (int i = 0; i < str.length(); i++) {
            if (!(str.charAt(i) == 'A' || str.charAt(i) == 'C' || str.charAt(i) == 'G' || str.charAt(i) == 'T')) {
                bool = false;
                break;
            }
        }
        return bool;
    }
}