LeetCode 刷题【187. 重复的DNA序列】

墨染点香2025-12-10 16:32

187. 重复的DNA序列

自己做

解：哈希

java 复制代码

class Solution {
    public List<String> findRepeatedDnaSequences(String s) {
        List<String> res = new ArrayList<>();
        int n = s.length();
        if (n <= 10) {
            return res;
        }

        // 哈希表：key=10长度子串，value=出现次数
        HashMap<String, Integer> countMap = new HashMap<>();

        // 遍历所有10长度子串，统计次数
        for (int i = 0; i <= n - 10; i++) {
            String sub = s.substring(i, i + 10);
            countMap.put(sub, countMap.getOrDefault(sub, 0) + 1);
        }

        // 收集出现次数≥2的子串
        for (String key : countMap.keySet()) {
            if (countMap.get(key) >= 2) {
                res.add(key);
            }
        }

        return res;
    }
}

看题解

java 复制代码

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

    public List<String> findRepeatedDnaSequences(String s) {
        List<String> ans = new ArrayList<String>();
        int n = s.length();
        if (n <= L) {
            return ans;
        }
        int x = 0;
        for (int i = 0; i < L - 1; ++i) {
            x = (x << 2) | bin.get(s.charAt(i));
        }
        Map<Integer, Integer> cnt = new HashMap<Integer, Integer>();
        for (int i = 0; i <= n - L; ++i) {
            x = ((x << 2) | bin.get(s.charAt(i + L - 1))) & ((1 << (L * 2)) - 1);
            cnt.put(x, cnt.getOrDefault(x, 0) + 1);
            if (cnt.get(x) == 2) {
                ans.add(s.substring(i, i + L));
            }
        }
        return ans;
    }
}

作者：力扣官方题解
链接：https://leetcode.cn/problems/repeated-dna-sequences/solutions/1035568/zhong-fu-de-dnaxu-lie-by-leetcode-soluti-z8zn/
来源：力扣（LeetCode）
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