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".

題意：要找出s中所有是p的變形字符串的子串，並返回子串起始位置。

解析：判斷一個子串是否是可以通過p變形得到，只需要判斷該子串所有字母出現次數與p是否一致，而題目中表明s和p都只含小寫字母，所以可以分別用一個大小爲26的數組去記錄子串和p中a~z出現的次數，然後比對兩個數組元素即可。然後只需要先對數組初始化，之後進行遍歷，每次遍歷子串往右移動一個位置，因此只需要在子串數組中修改移除的以及新加入的字母對應數組元素大小，然後每次進行判斷兩個數組元素是否一致即可。算法複雜度爲O(s.length)，具體代碼如下：

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;
	}
};