LeetCode 438. 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".

 

这道题要求统计原字符串中，与模板字符串构成Anagrams的子串有多少，所谓Anagrams，也就是在寻找pattern的一个全排列，但是这里不用这么麻烦，只需要统计patter中各个字母的出现次数，之后逐步滑动，判断原字符串中与子串等长的字符串个字母出现的次数，patter和源字符串统计结果相等，那就说明该位置匹配

class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        vector<int> sv(26, 0), pv(26, 0), res;
        if (s.size() < p.size())
            return res;
        for (int i =  0; i < p.size(); i++)
        {
            pv[p[i]-'a']++;
            sv[s[i]-'a']++;
        }
        if (sv == pv)
            res.push_back(0);
        for (int i = p.size(); i < s.size(); i++)
        {
            sv[s[i]-'a']++;
            sv[s[i-p.size()]-'a']--;
            if (pv == sv)
                res.push_back(i - p.size() + 1);
        }
        return res;
    }
};

当然也可以做一些优化思路：

class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        if (s.empty()) return {};
        vector<int> res, m(256, 0);
        int left = 0, right = 0, cnt = p.size(), n = s.size();
        for (char c : p) ++m[c];
        while (right < n) {
            if (m[s[right++]]-- >= 1) --cnt;
            if (cnt == 0) res.push_back(left);
            if (right - left == p.size() && m[s[left++]]++ >= 0) ++cnt;
        }
        return res;
    }
};

这种方法只需要遍历一遍并且不需要比较两个向量的大小