187. Repeated DNA Sequences重复的DNA子串序列

[抄题]:

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.

Example:

Input: s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"

Output: ["AAAAACCCCC", "CCCCCAAAAA"]

 [暴力解法]:

时间分析:

空间分析:

 [优化后]:

时间分析:

空间分析:

[奇葩输出条件]:

  1. set转成arraylist直接放到括号里就行了

[奇葩corner case]:

[思维问题]:

[英文数据结构或算法,为什么不用别的数据结构或算法]:

10个数,从开始算,加到9就可以了。

for (int i = 0; i + 9 < s.length(); i++)

.substring包左不包右,所以必须写十位数。但是inde

String ten = s.substring(i, i + 10);
 
beginIndex =< str的值 < endIndex

[一句话思路]:

[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):

[画图]:

[一刷]:

  1. set的判断语句 没加就自己自动加 没必要再写一遍
  2. set用add,map用put

[二刷]:

[三刷]:

[四刷]:

[五刷]:

  [五分钟肉眼debug的结果]:

[总结]:

set就是判重,一个不够用可以用两个

[复杂度]:Time complexity: O(n) Space complexity: O(n)

[算法思想:迭代/递归/分治/贪心]:

[关键模板化代码]:

[其他解法]:

[Follow Up]:

[LC给出的题目变变变]:

 [代码风格] :

 [是否头一次写此类driver funcion的代码] :

 [潜台词] :

class Solution {
    public List<String> findRepeatedDnaSequences(String s) { 
        //ini: two sets
        List<String> result = new ArrayList<String>();
        Set<String> seen = new HashSet<String>();
        Set<String> ten = new HashSet<String>();
        
        //cc
        if (s.length() == 0) return result;
        
        //for loop: get substring
        for (int i = 0; i + 9 < s.length(); i++) {
            String str = s.substring(i, i+ 10);
            if (!seen.add(str)) ten.add(str);
        }
        
        //return
        return new ArrayList(ten);
    }
}
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