Find All Anagrams in a String
来源:互联网 发布:万方数据库和cnki差别 编辑:程序博客网 时间:2024/05/30 04:35
题目地址:https://leetcode.com/problems/find-all-anagrams-in-a-string/
Given a string s and a non-empty string p, find all the start indices of p’s anagrams in s.
Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.
The order of output does not matter.
Example 1:
Input:s: "cbaebabacd" p: "abc"Output:[0, 6]Explanation:The substring with start index = 0 is "cba", which is an anagram of "abc".The substring with start index = 6 is "bac", which is an anagram of "abc".
Example 2:
Input:s: "abab" p: "ab"Output:[0, 1, 2]Explanation:The substring with start index = 0 is "ab", which is an anagram of "ab".The substring with start index = 1 is "ba", which is an anagram of "ab".The substring with start index = 2 is "ab", which is an anagram of "ab".
这个题目的意思就是说在字符串s中,是否存在这样的子字符串,这个子字符串是字符串p的一种排列,如果存在,这输出这样的子字符串的所有的位置。
首先假如说s中存在这样的子字符串substr,那么substr的长度肯定跟p的长度是相等的。下来就是比较substr与p的关系了,是不是互为一种排列。
确定这种关系有很多种,比如将p和substr都排个序,然后比较排个序的两个字符串是不是相等,相等则互为排列,不相等则不互为排列。
或者用p中的每一个字符与substr中的字符比较,看是否存在,如果存在,则移除,注意:这一步移除很重要,如果不移除,很可能会重复比较,比如“ab”和“aa”,继续比较,直到p中的字符遍历完。如果不存在,那么就不再往下比较了,继续进行下一轮迭代。
实现代码如下:
public class FindAllAnagramsInAString { public List<Integer> findAnagrams(String s, String p) { List<Integer> result = new ArrayList<Integer>(); int pLen = p.length(); for (int i = 0; i <= s.length() - pLen; i++) { String subStr = s.substring(i, i + pLen); boolean isTrue = true; for (int j = 0; j < pLen; j++) { if (subStr.indexOf(p.charAt(j)) == -1) { isTrue = false; break; } else { subStr = subStr.substring(0, subStr.indexOf(p.charAt(j))) + subStr.substring(subStr.indexOf(p.charAt(j)) + 1); continue; } } if (isTrue) result.add(i); } return result; } public static void main(String[] args) { FindAllAnagramsInAString findAllAnagramsInAString = new FindAllAnagramsInAString(); System.out.println(findAllAnagramsInAString.findAnagrams("baa", "aa")); }}
以上实现虽然实现了相应的功能,但是时间复杂度比较高,不能通过所有的test case,主要问题还是有些案例超时,所以还需要进一步优化算法。
今天就写到这里,后续更新。
- 438. Find All Anagrams in a String
- 438. Find All Anagrams in a String
- LeetCode Find All Anagrams in a String
- 438. Find All Anagrams in a String
- leetcode: Find All Anagrams in a String
- leetcode438. Find All Anagrams in a String
- leetcode438. Find All Anagrams in a String
- 438. Find All Anagrams in a String
- 438. Find All Anagrams in a String
- [LeetCode]Find All Anagrams in a String
- Find All Anagrams in a String
- LeetCode438 Find All Anagrams in a String
- 438. Find All Anagrams in a String
- Leetcode Find All Anagrams in a String
- 438. Find All Anagrams in a String
- 438. Find All Anagrams in a String
- 438. Find All Anagrams in a String*
- 438. Find All Anagrams in a String
- 得到当期日期测试
- 学习java哪些东西必须牢记
- [selenium+python]环境
- C语言数组与指针详解
- 约瑟夫环的递推方法
- Find All Anagrams in a String
- zookeeper中的自动交互引发的思考
- git基本操作从远程服务器获取项目
- Servlet
- 同义词,近义词,反义词收集
- 《拖延心理学》
- jQuery -原生 如何互转
- java语言学习是否必须英语跟数学有一定基础
- Slow Boot Slow Logon (SBSL), A Tool Called XPerf and Links You Need To Read