标签：tps i++ span order nta key color output 分组
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 pwill 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".
根据滑动窗口的思想，利用map记录count，时间复杂度为O(n)
public List<Integer> findAnagrams(String s, String p) {//map 滑动窗口 mytip List<Integer> re = new ArrayList<>(); if(null==s||null==p||s.length()<p.length()){ return re; } Map<Character,Integer> map = new HashMap<>(); for (int i = 0; i < p.length(); i++) { char c= p.charAt(i); if(map.containsKey(c)){ map.put(c,map.get(c)+1); } else{ map.put(c,1); } } int left =0; int right =0; int count = p.length(); while(right<s.length()){ char c = s.charAt(right); right++; if(map.containsKey(c)){ if(map.get(c)>=1){ count--; } map.put(c,map.get(c)-1); } if(0==count){ re.add(left); } if(right-left==p.length()){ char cl = s.charAt(left); left++; if(map.containsKey(cl)){ if(0<=map.get(cl)){ count++; } map.put(cl,map.get(cl)+1); } } } return re; }
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标签：tps i++ span order nta key color output 分组
原文地址：https://www.cnblogs.com/zhacai/p/10596274.html
