【LeetCode】Repeated DNA Sequences 解题报告

【题目】

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"].

【解析】

题意：找出DNA序列中所有出现次数大于1的长度为10的子串。

我觉得题目给的例子有问题，例子中除了给出的两个子串，还有“ACCCCCAAAA", "AACCCCCAAA", "AAACCCCCAA", "AAAACCCCCA"也都符合条件。

参考 https://oj.leetcode.com/discuss/24478/i-did-it-in-10-lines-of-c 我来解释一下思路：

首先看这道题的Tags是Hash Table和Bit Manipulation，那么如何把字符串转化为位操作呢？我们首先来看字母 ”A" "C" “G" "T" 的ASCII码，分别是65, 67, 71, 84，二进制表示为 1000001, 1000011, 1000111, 1010100。可以看到它们的后四位是不同，所以用后四位就可以区分这四个字母。一个字母用3bit来区分，那么10个字母用30bit就够了。用int的第29~0位分表表示这0~9个字符，然后把30bit转化为int作为这个子串的key，放入到HashTable中，以判断该子串是否出现过。

【Java代码】

public class Solution {
    public List<String> findRepeatedDnaSequences(String s) {
        List<String> ans = new ArrayList<String>();
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        int key = 0;
        for (int i = 0; i < s.length(); i++) {
            key = ((key << 3) | (s.charAt(i) & 0x7)) & 0x3fffffff;
            if (i < 9) continue;
            if (map.get(key) == null) {
                map.put(key, 1);
            } else if (map.get(key) == 1) {
                ans.add(s.substring(i - 9, i + 1));
                map.put(key, 2);
            }
        }
        return ans;
    }
}

【补充】

如果说不记得A，C，G，T的ASCII码了，而且也不想计算它们的二进制表示，那么可以用一种替代的方法，即把A，C，G，T映射成0，1，2，3，这样用两个bit就可以区分它们00，01，10，11了。还有这里是长度为10的子串，而且这四个字符用3bit就可以区分，加起来总共30bit，如果子串长度再长些，或者字符种类再多些需要3bit以上来区分，总共bit数超过32，那么采用映射不失为一种很好的解决方法。可参见http://bookshadow.com/weblog/2015/02/06/leetcode-repeated-dna-sequences/

时间： 2024-10-07 09:47:43

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