leetcode——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".

题意：要找出s中所有是p的变形字符串的子串，并返回子串起始位置。

解析：判断一个子串是否是可以通过p变形得到，只需要判断该子串所有字母出现次数与p是否一致，而题目中表明s和p都只含小写字母，所以可以分别用一个大小为26的数组去记录子串和p中a~z出现的次数，然后比对两个数组元素即可。然后只需要先对数组初始化，之后进行遍历，每次遍历子串往右移动一个位置，因此只需要在子串数组中修改移除的以及新加入的字母对应数组元素大小，然后每次进行判断两个数组元素是否一致即可。算法复杂度为O(s.length)，具体代码如下：

class Solution {public:vector<int> findAnagrams(string s, string p) {vector<int> ans;int ls = s.length(), lp = p.length();int ns[26] = {0}, np[26] = {0};for(int i=0; i<lp; i++){np[p[i]-'a'] ++;ns[s[i]-'a'] ++;}if(Anagrams(np, ns))ans.push_back(0);for(int i=1; i<=ls-lp; i++){ns[s[i-1]-'a'] --;ns[s[i+lp-1]-'a'] ++;if(Anagrams(np, ns))ans.push_back(i);}return ans;}bool Anagrams(int a[], int b[]){for(int i=0; i<26; i++)if(a[i] != b[i])return false;return true;}};