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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;
}
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