[LeetCode]Find All Anagrams in a String

niceforbear 發(fā)布于2019-08-16 10:26 / 3074人閱讀

摘要：解題思路，就是只順序不同但個(gè)數(shù)相同的字符串，那我們就可以利用的思想來比較每個(gè)字符串中字符出現(xiàn)的個(gè)數(shù)是否相等。

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.解題思路

anagrams，就是只順序不同但個(gè)數(shù)相同的字符串，那我們就可以利用hashtable的思想來比較每個(gè)字符串中字符出現(xiàn)的個(gè)數(shù)是否相等。
對于兩個(gè)字符串我們分別準(zhǔn)備數(shù)組(大小為256)來存儲(chǔ)每個(gè)字符出現(xiàn)的次數(shù)：
1) 對于p，我們遍歷，并在hp中記錄字符出現(xiàn)的次數(shù)；
2) 之后遍歷s,先把當(dāng)前字符的個(gè)數(shù)+1，但是需要考慮當(dāng)前index是否已經(jīng)超過了p的長度，如果超過，則表示前面的字符已經(jīng)不予考慮，所以要將index-plen的字符的個(gè)數(shù)-1；最后判斷兩個(gè)數(shù)組是否相等，如果相等，返回index-plen+1,即為開始的下標(biāo)。

2.代碼

public class Solution {
    public List findAnagrams(String s, String p) {
        List res=new ArrayList();
        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){
                hs[s.charAt(j-plen)]--;
            }
            if(Arrays.equals(hs,hp))
                res.add(j-plen+1);
        }
        return res;
    }
}