import java.util.ArrayList;

import java.util.List;

/**

* created by zl on 2019/3/30 10:11

*/

public class simple_438_FindAllAnagramsinaString

{

public List<Integer> findAnagrams(String s, String p)

{

List<Integer> res = new ArrayList<>();

int m=s.length();

int n=p.length();

if(m<n)

return res;

int[] map = new int[26];

for(char ch: p.toCharArray())

map[ch-'a']++; //how many letters of each kind in the mode p

int counter=n;

int left=0, right=0; //两个指针表示窗口的左端和右端

char []ss = s.toCharArray();

while(right<m)

{

System.out.println("map[ss[right]-'a']="+map[ss[right]-'a']+", ss[right]="+ss[right]+", right="+right);

if(map[ss[right]-'a']-- >=1) //there's a letter of mode

counter--;

right++; //每次都要向右移动一次, 如果对应字符出现次数大于等于1说明该字符在p中, 将counter减1

if(counter==0) // 如果counter减为0, 将左端点index加入返回值

res.add(left);

System.out.println("right="+right+",count="+counter);

if(right-left==n)

{

//说明窗口大小超出限制, 需要left右移一位同时判断left对应字符是否在p中,并决定是否对counter加1

System.out.println("map[ss[left]-'a']="+map[ss[left]-'a']+", ss[left]="+ss[left]+", left="+left);

if(map[ss[left]-'a']++>=0)

counter++;

left++;

}

}

return res;

}

public static void main(String[] args) {

simple_438_FindAllAnagramsinaString findAllAnagramsinaString = new simple_438_FindAllAnagramsinaString();

findAllAnagramsinaString.findAnagrams("cbaebabacd","abc");

}

}