LeetCode 187. Repeated DNA Sequences

最新推荐文章于 2023-03-18 22:02:57 发布

ActiveCoder

最新推荐文章于 2023-03-18 22:02:57 发布

阅读量179

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode 题解

本文链接：https://blog.youkuaiyun.com/github_34333284/article/details/51433850

LeetCode 题解专栏收录该内容

288 篇文章

订阅专栏

本文介绍了一个函数，用于在DNA分子中查找所有长度为10的重复序列子串。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

#include <unordered_map>
#include <vector>
#include <iostream>
using namespace std;

/*
   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.

  Given s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT",

  Return:
  ["AAAAACCCCC", "CCCCCAAAAA"].
*/

vector<string> findRepeatedDnaSequences(string s) {
  if(s.size() < 10) return {};
  vector<string> res;
  unordered_map<string, int> indexToString;
  for(int i = 0; i <= s.size() - 10; ++i) {
    string tmp = s.substr(i, 10);
    auto iter = indexToString.find(tmp);
    if(iter == indexToString.end()) indexToString.insert({tmp, 1});
    else iter->second += 1;
  }
  auto iter = indexToString.begin();
  while(iter != indexToString.end()) {
    if(iter->second > 1) res.push_back(iter->first);
    iter++;
  }
  return res;
}

int main(void) {
  vector<string> res = findRepeatedDnaSequences("AAAAAAAAAAA");
  for(int i = 0; i < res.size(); ++i) {
    cout << res[i] << endl;
  }
}