438. Find All Anagrams in a String
作者:互联网
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".
public class Solution { public List<Integer> findAnagrams(String s, String t) { List<Integer> result = new LinkedList<>(); if(t.length()> s.length()) return result; Map<Character, Integer> map = new HashMap<>(); for(char c : t.toCharArray()){ map.put(c, map.getOrDefault(c, 0) + 1); } int counter = map.size(); int begin = 0, end = 0; int head = 0; int len = Integer.MAX_VALUE; while(end < s.length()){ char c = s.charAt(end); if( map.containsKey(c) ){ map.put(c, map.get(c)-1); if(map.get(c) == 0) counter--; System.out.print(counter + " "); System.out.println(c + " " + map.get(c)); } end++; while(counter == 0){ char tempc = s.charAt(begin); if(map.containsKey(tempc)){ map.put(tempc, map.get(tempc) + 1); if(map.get(tempc) > 0){ counter++; } } if(end-begin == t.length()){ result.add(begin); } System.out.println("begin " + begin); begin++; } } return result; } }
标签:map,begin,ab,Anagrams,get,substring,start,438,Find 来源: https://www.cnblogs.com/wentiliangkaihua/p/12908859.html