leetcode oj java Find All Anagrams in a String

一、问题描述：

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".

二、思路：

      p相当于一个滑块，每次往后滑一次。 在比较的时候用到一个128长度的数组记录每个char出现的次数，判断p和滑到的区间中字符出现的次数是否一致即可。

三、代码：

public class Solution {    public boolean compare(int[] v1, int[] v2) {        boolean re = true;        for (int i = 0; i < v1.length; i++) {            if (v1[i] != v2[i]) {                return false;            }        }        return re;    }    public List<Integer> findAnagrams(String s, String p) {         List<Integer> re = new ArrayList<Integer>();        int[] pNum = new int[128];        for (int i = 0; i < p.length(); i++) {            pNum[p.charAt(i)]++;        }        for (int j = 0; j < s.length() - p.length() + 1; j++) {            boolean flag = true;            if (pNum[s.charAt(j)] > 0) {                String tmp = s.substring(j, j + p.length());                int[] sNum = new int[128];                for (int i = 0; i < tmp.length(); i++) {                    if (pNum[s.charAt(j)] == 0) {                        flag = false;                        break;                    } else {                        sNum[tmp.charAt(i)]++;                    }                }                if (flag && compare(sNum, pNum)) {                    re.add(j);                }            }        }        return re;    }}