LeetCode OJ Repeated DNA Sequences

最新推荐文章于 2021-10-10 00:59:49 发布

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本文介绍了一种通过将DNA序列转换为数字的方法来查找长度为10的重复子序列。利用布尔数组跟踪已出现过的序列，并在发现重复序列时将其记录下来。

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"].

4^10=1048576, 可将DNA序列转为数字来做。

bool nums[1048576], isIn[1048576];
class Solution {
public:
	vector<string> findRepeatedDnaSequences(string s) {
		int size = s.length(), sum = 0;
		char letter[4] = {'A', 'C', 'G', 'T'};
		memset(nums, false, sizeof(nums));
		memset(isIn, false, sizeof(isIn));
		vector<string> ans;
		for (int i = 0; i < 9; i++) {
			if (s[i] == 'A') sum = sum * 4 + 0;
			if (s[i] == 'C') sum = sum * 4 + 1;
			if (s[i] == 'G') sum = sum * 4 + 2;
			if (s[i] == 'T') sum = sum * 4 + 3;
		}
		for (int i = 9; i < size; i++) {
			if (s[i] == 'A') sum = sum * 4 + 0;
			if (s[i] == 'C') sum = sum * 4 + 1;
			if (s[i] == 'G') sum = sum * 4 + 2;
			if (s[i] == 'T') sum = sum * 4 + 3;
			sum %= 1048576;
			if (isIn[sum]) continue;
			else if (nums[sum]) {
				isIn[sum] = true;
				string in;
				int S = sum;
				for (int j = 0; j < 10; j++) {
					in.push_back(letter[S % 4]);
					S /= 4;
				}
				reverse(in.begin(), in.end());
				ans.push_back(in);
			}
			nums[sum] = true;
		}
		return ans;
	}
};