438. Find All Anagrams in a String
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2022-07-12 08:38:14
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原题
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".
代码实现
这个算法的时间复杂度为O(n),空间复杂度为O(1)。这个算法比较难理解,以后慢慢回头琢磨吧。
public IList<int> FindAnagrams(string s, string p){
IList<int> rtn = new List<int>();
int[] hash = new int[123]; //a~z: 97~122
foreach (var c in p){
hash[Convert.ToInt32(c)]++; //hash: key:char, value: occuring times
}
int eachBeg = 0, eachEnd = 0, count = p.Length;
while (eachEnd < s.Length){
char tmpchar = s[eachEnd];
if (hash[tmpchar] >= 1)
count--;
hash[tmpchar]--;
eachEnd++; //every time the eachEnd pointer to move toward right
if (count == 0)
rtn.Add(eachBeg);
//reset the hash.
if (eachEnd - eachBeg == p.Length){
char tmp = s[eachBeg];
if (hash[tmp] >= 0)
count++;
hash[tmp]++;
eachBeg++;
}
}
return rtn;
}
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