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出处 LeetCode 算法第187题 所有 DNA 由一系列缩写为 A,C,G 和 T 的核苷酸组成,例如:“ACGAATTCCG”。在研究 DNA 时,识别 DNA 中的重复序列有时会对研究非常有帮助。 编写一个函数来查找 DNA 分子中所有出现超多一次的10个字母长的序列(子串)。 示例: 输入: s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT" 输出: ["AAAAACCCCC", "CCCCCAAAAA"]
出处 LeetCode 算法第187题
所有 DNA 由一系列缩写为 A,C,G 和 T 的核苷酸组成,例如:“ACGAATTCCG”。在研究 DNA 时,识别 DNA 中的重复序列有时会对研究非常有帮助。
编写一个函数来查找 DNA 分子中所有出现超多一次的10个字母长的序列(子串)。
示例:
输入: s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"
输出: ["AAAAACCCCC", "CCCCCAAAAA"]
遍历所有10个字符的组合,并将其当做key存于map中,找出其中value对于1的即可
JavaScript
var findRepeatedDnaSequences = function (s) { var length = s.length, map = {}, result = []; for (var i = 0; i <= length - 10; i++) { var temp = s.slice(i, i + 10); if (map[temp]) { map[temp]++; } else { map[temp] = 1; } } for (var key in map) { if (map[key] > 1) { result.push(key); } } return result; };
Java
class Solution { public List<String> findRepeatedDnaSequences(String s) { int length = s.length(); Map<String, Integer> map = new HashMap<>(); List<String> result = new ArrayList<>(); for (int i = 0; i <= length - 10; i++) { String temp = s.substring(i, i + 10); if (map.containsKey(temp)) { map.put(temp, map.get(temp) + 1); } else { map.put(temp, 1); } } for (String key : map.keySet()) { if (map.get(key) > 1) { result.add(key); } } return result; } }
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习题
思路
遍历所有10个字符的组合,并将其当做key存于map中,找出其中value对于1的即可
解答
JavaScript
Java
The text was updated successfully, but these errors were encountered: