438 .Find All Anagrams in a String

438 .Find All Anagrams in a String

题目描述：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".

题目大意:给定两个字符串s与p，找出s中出现p的起始位置

思路：此类题通用方法 https://discuss.leetcode.com/topic/68976/sliding-window-algorithm-template-to-solve-all-the-leetcode-substring-search-problem/2

代码

package HashTable;

import java.util.*;

/**
* @author OovEver
* 2018/1/11 14:32
*/
public class LeetCode438 {
public static 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--;
}
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);
}
begin++;
}

}
return result;
}

}