[LeetCode]Find All Anagrams in a String

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".

1.解题思路

anagrams，就是只顺序不同但个数相同的字符串，那我们就可以利用hashtable的思想来比较每个字符串中字符出现的个数是否相等。
对于两个字符串我们分别准备数组(大小为256)来存储每个字符出现的次数：
1) 对于p，我们遍历，并在hp中记录字符出现的次数；
2) 之后遍历s,先把当前字符的个数+1，但是需要考虑当前index是否已经超过了p的长度，如果超过，则表示前面的字符已经不予考虑，所以要将index-plen的字符的个数-1；最后判断两个数组是否相等，如果相等，返回index-plen+1,即为开始的下标。

2.代码

public class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        List<Integer> res=new ArrayList<Integer>();
        if(s.length()==0||s==null||p.length()==0||p==null) return res;
        int[] hs=new int[256];
        int[] hp=new int[256];
        int plen=p.length();
        for(int i=0;i<plen;i++){
            hp[p.charAt(i)]++;
        }
        for(int j=0;j<s.length();j++){
            hs[s.charAt(j)]++;
            if(j>=plen){
                hs[s.charAt(j-plen)]--;
            }
            if(Arrays.equals(hs,hp))
                res.add(j-plen+1);
        }
        return res;
    }
}