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  • 找出字符串里所有的Anagrams(Find All Anagrams in a String)

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    题目:

    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.

    题意为给定一个字符串s和一个非空字符串p,找出s中所有是p的anagrams的子串的起始索引,所有字符串均由小写字母组成,且字符串长度不超过20100,将结果以列表形式返回,不必考虑列表的顺序。感兴趣的童鞋可以戳这里看原题

    例如:

    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".
    

    算法:

    先上代码,这里用python3实现:

    class Solution:
        def findAnagrams(self, s, p):
            """
            :type s: str
            :type p: str
            :rtype: List[int]
            """
            res=[]
            m,n=len(s),len(p)
            phash=[0]*123
            shash=[0]*123
            if n>m:return res
            for i in p:
                phash[ord(i)]+=1
            for i in s[:n-1]:
                shash[ord(i)]+=1
            for i in range(n-1,m):
                shash[ord(s[i])]+=1
                if i-n>=0:
                    shash[ord(s[i-n])]-=1
                if shash==phash:
                    res.append(i-n+1)
            return res
    

    思路是这样的,考虑到字符串由小写字母组成,所以可利用哈希表和ascii编码,字母的编码范围是65~122,所以可以创建连个长度为123键值全为0的哈希表,先对p遍历,将p中出现的字母及次数用哈希表phash记录下来,再依次遍历s的每个字母并记录到shash,并与phash进行比对,若相等,则说明找到了一个anagrams,将此时的起始索引存进res列表即可,直到s遍历完,返回res。

    初心易得,始终难守。
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  • 原文地址:https://www.cnblogs.com/Aurora-Twinkle/p/8688129.html
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