[LeetCode] 438. Find All Anagrams in a String

轉自LeetCode

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".

<Solution>

這題是 242. Valid Anagram 的進階題

解題想法如下

• 一樣使用 hash map 來當作查表

• 利用兩個指標，來做一個 window 的概念，歷遍 s

• 當 p 裡面的字元都用光的時候，檢查是否符合條件，並且更新狀態

這類的題目有一個樣版，可參考這篇文章

code 如下

Java

	class Solution {
	public List<Integer> findAnagrams(String s, String p) {
	//>> initialize
	List<Integer> ans = new LinkedList<>();
	if(p.length() > s.length()) {
	return ans;
	}
	//>> hash
	int[] map = new int[26];
	for(final char c : p.toCharArray()) {
	map[c-'a']++;
	}
	//>> maintain a counter
	final int targetLen = p.length();
	int count = targetLen;
	//>> loop from the begining of the string
	int start = 0, end = 0;
	final int inputLen = s.length();
	char c;
	while(end < inputLen) {
	//>> check hash and modify the counter according to the requirement
	c = s.charAt(end++);
	if(map[c-'a']-- > 0) {
	--count;
	}
	//>> counter condition
	while(count == 0) {
	//>> modify the counter if it is necessary
	c = s.charAt(start);
	if(++map[c-'a'] > 0) {
	++count;
	}
	//>> check the answer according to the requirement
	if(end-start == targetLen) {
	ans.add(start);
	}
	//>> increase begin pointer to make it invalid/valid again
	++start;
	}
	}
	//>> return ans
	return ans;
	}
	}

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