[Leetcode] 438. Find All Anagrams in a String 解题报告

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.

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".

思路：

一道简单的题目。下面的代码我参考了小榕流光的，他用v
4000
ector<int>数组来模拟哈希表，比较巧妙。比较hash1和hash2是否相等需要O(256)的时间复杂度，可能会导致算法时间复杂度的常系数比较大。

代码：

class Solution {
public:
vector<int> findAnagrams(string s, string p) {
vector<int> hash1(256, 0), hash2(256, 0);
for(auto ch: p) {
++hash1[ch];
}
int lenp = p.size(), lens = s.size();
vector<int> ans;
for(int i = 0; i < lens; i++) {
++hash2[s[i]];
if(i >= lenp) {
hash2[s[i-lenp]]--;
}
if(hash1 == hash2) {
ans.push_back(i- lenp + 1);
}
}
return ans;
}
};