Leetcode 438 Find All Anagrams in a String

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".

这题我真是做复杂了。。我自己的做法惨不忍睹 虽然也是window的基本思想然而太复杂了

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

ArrayList<Integer> result = new ArrayList<Integer>();
int[] hash = new int[256];
for (char c : p.toCharArray()) {
hash[c]=0;}

int[] example = new int[256];
for (char c : p.toCharArray()) {
example[c]++;}

char[] str = s.toCharArray();
char[] ptr = p.toCharArray();

for(int i =0;i<str.length-p.length()+1;i++){
// System.out.println(i);
int flag = 0;
for(int j=0;j<p.length();j++){

hash[str[i+j]]++;

if(j==p.length()-1){
for(int m=0;m<p.length();m++){
// System.out.println(ptr[m]+"="+hash[ptr[m]]);
if(hash[ptr[m]]!=example[ptr[m]]) flag= 1;
}
if(flag==0){
result.add(i);
// System.out.println("lalala");
}

// System.out.println("clear");
for (char c : p.toCharArray()) {
hash[c]=0;}
}
}

}
return result;
}

public List<Integer> findAnagrams(String s, String p) {
List<Integer> list = new ArrayList<>();
if (s == null || s.length() == 0 || p == null || p.length() == 0) return list;//首先要排除特殊情况
int[] hash = new int[256];

for (char c : p.toCharArray()) {
hash[c]++;
}

int left = 0, right = 0, count = p.length();//这里使用了双指针 作为window的前后端
while (right < s.length()) {//使用window的右端是否出界来判断循环是否

if (hash[s.charAt(right++)]-- >= 1) count--;

//when the count is down to 0, means we found the right anagram
//then add window's left to result list
if (count == 0) list.add(left);

//if we find the window's size equals to p, then we have to move left (narrow the window) to find the new match window
//++ to reset the hash because we kicked out the left
//only increase the count if the character is in p
//the count >= 0 indicate it was original in the hash, cuz it won't go below 0
if (right - left == p.length() && hash[s.charAt(left++)]++ >= 0) count++;
}
return list;
}