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".

滑動視窗，每次新增右邊的一個字元，刪除左邊的一個字元，看看是不是滿足各種字元數都大於等於p中各種字元的字元數的條件

注意檢測的時候完全沒必要掃描窗口裡的所有字元來統計個數，視窗滑動時只需要考慮一個新增進來的字元和一個刪除的字元，因為只有這兩個字元出現了變動，用needed標記“缺失”的字元數，假設字元為X，串P中字元X的個數為P(X)，當X等於P(X)且X需要刪除的時候才會使得“缺失”的字元數+1 ；當X等於P(X)-1且需要新增的時候才會使得“缺失”的字元數-1

public class Solution {
    public List<Integer> findAnagrams(String s, String p)
	{
		List<Integer> retlist=new ArrayList<>();
		int slen=s.length();
		int plen=p.length();
		if(slen<plen)
			return retlist;
		int[] carr=new int[128];
		for(int i=0;i<plen;i++)
			carr[p.charAt(i)]++;
		
		int need=0;
		int[] count=new int[128];
		
		for(int i=0;i<plen;i++)
			count[s.charAt(i)]++;
		
		for(char c='a';c<='z';c++)
			if(count[c]<carr[c])
				need++;
		
		if(need==0)
			retlist.add(0);
		
		for(int i=plen;i<slen;i++)
		{
			char preChar=s.charAt(i-plen);
			char nowChar=s.charAt(i);
			if(--count[preChar]==carr[preChar]-1)
				need++;
			if(++count[nowChar]==carr[nowChar])
				need--;
			if(need==0)
				retlist.add(i-plen+1);
		}
		
		return retlist;
	}
}