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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> leetcode 187: Repeated DNA Sequences

leetcode 187: Repeated DNA Sequences

編輯:C++入門知識

leetcode 187: Repeated DNA Sequences


Total Accepted: 1161 Total Submissions: 6887

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"].

[分析]

HASHMAP方法會EXCEED SPACE LIMIT.

因為只有4個字母,所以可以創建自己的hashkey, 每兩個BITS, 對應一個 incoming character. 超過20bit 即10個字符時, 只保留20bits.

[注意]

1. (hash<<2) + map.get(c) 符號優先級, << 一定要括起來.


public class Solution {
    public List findRepeatedDnaSequences(String s) {
        List res = new ArrayList();
        if(s==null || s.length() < 11) return res;
        int hash = 0;
        
        Map map = new HashMap();
        map.put('A', 0);
        map.put('C', 1);
        map.put('G', 2);
        map.put('T', 3);
        
        Set set = new HashSet();
        Set unique = new HashSet();
        
        for(int i=0; i

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