LC.P187[重复的DNA序列]

方法一：滑动窗口+哈希表

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class Solution {
    public List<String> findRepeatedDnaSequences(String s) {
        List<String> ans = new ArrayList<>();
        Map<String, Integer> map = new HashMap<>();
        for (int i = 0; i + 10 <= s.length(); ++i) {
            String cur = s.substring(i, i + 10);
            if (map.merge(cur, 1, Integer::sum) == 2) ans.add(cur);
        }
        return ans;
    }
}
  • 时间复杂度：每次检查以s[i]为结尾的子串，需要构造出新的长度为$10$的字符串。令$C = 10$，复杂度为$O(n \times C)$
  • 空间复杂度：$O(n)$

方法二：滑动窗口+哈希表+位运算

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class Solution {
    static int length = 10;
    Map<Character, Integer> map = new HashMap<>() {{
        put('A', 0);
        put('C', 1);
        put('G', 2);
        put('T', 3);
    }};

    public List<String> findRepeatedDnaSequences(String s) {
        List<String> ans = new ArrayList<>();
        int n = s.length();
        if (n <= length) return ans;
        int x = 0;
        for (int i = 0; i < length - 1; ++i) {
            x = (x << 2) | map.get(s.charAt(i));
        }
        Map<Integer, Integer> cnt = new HashMap<>();
        for (int i = 0; i + length <= n; ++i) {
            x = ((x << 2) | map.get(s.charAt(i + length - 1))) & ((1 << (length * 2)) - 1);
            cnt.put(x, cnt.getOrDefault(x, 0) + 1);
            if (cnt.get(x) == 2) {
                ans.add(s.substring(i, i + length));
            }
        }
        return ans;
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$