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博客园 - NickyYe

DYNAMIC LINK LIBRARY - DLL 分布式系统中Unique ID 的生成方法 205. Isomorphic Strings 201. Bitwise AND of Numbers Range 189. Rotate Array 167. Two Sum II - Input array is sorted Convert BST to Greater Tree Uncommon Words from Two Sentences Path Sum III Delete Node in a BST Sliding Window Maximum Find K Closest Elements C++ TUTORIAL - MEMORY ALLOCATION - 2016 多线程 Console Event Handling SetConsoleCtrlHandler() -- 设置控制台信号处理函数 SetConsoleCtrlHandler 处理控制台消息 总结open与fopen的区别 LevelDB
187. Repeated DNA Sequences
NickyYe · 2018-12-24 · via 博客园 - NickyYe

https://leetcode.com/problems/repeated-dna-sequences/

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.

Example:

Input: s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"

Output: ["AAAAACCCCC", "CCCCCAAAAA"]

解题思路:

需要注意的是,因为substring的长度是const的10,所以实际上是O(10n)。

class Solution {
    public List<String> findRepeatedDnaSequences(String s) {
        List<String> res = new ArrayList<String>();
        
        if (s.length() < 11) {
            return res;
        }
        
        StringBuffer sb = new StringBuffer();
        Map<String, Integer> map = new HashMap<String, Integer>();        
        int index = 0;
        
        while (index <= s.length() - 10) {
            String temp = s.substring(index, index + 10);
            int count = map.getOrDefault(temp, 0);
            if (count == 1) {
                res.add(temp);
                map.put(temp, count + 1);
            } else if (count == 0){
                map.put(temp, 1);
            }
            index++;
        }
        
        return res;
    }
}