【LeetCode】Find All Anagrams in a String

最新推荐文章于 2020-05-18 00:23:42 发布

原创最新推荐文章于 2020-05-18 00:23:42 发布 · 159 阅读

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本文介绍了一种算法，用于在给定的字符串s中找到所有字符串p的变位词的起始索引。通过比较两个字符串中字符的频率，可以高效地检测出变位词。文章提供了一个C++实现的例子。

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题目描述：

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".

题目大意：找出字符串中的变位词，变位词指的是，字符种类、个数均相同但顺序不同的两个词

所以我们必须先算出字符串p中字符出现的次数，然后从s的头开始找p的长度的字符，验证字符个数是否相同

不同就直接break，相同的话就将起始位置加入到结果中

class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        if (s.empty()) return {};
        vector<int> res, m1(256, 0), m2(256, 0);
        for (int i = 0; i < p.size(); ++i) {
            ++m1[s[i]]; ++m2[p[i]];
        }
        if (m1 == m2) res.push_back(0);
        for (int i = p.size(); i < s.size(); ++i) {
            ++m1[s[i]]; 
            --m1[s[i - p.size()]];
            if (m1 == m2) res.push_back(i - p.size() + 1);
        }
        return res;
    }
};