[LeetCode]Find All Anagrams in a String

niceforbear 发布于2019-08-16 10:26 / 2901人阅读

摘要：解题思路，就是只顺序不同但个数相同的字符串，那我们就可以利用的思想来比较每个字符串中字符出现的个数是否相等。

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 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;
    }
}

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