LeetCode 438. Find All Anagrams in a String （在字符串中找到所有的变位词）

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".

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题目标签：Hash Table

　　题目给了我们两个string s 和 p， 让我们在 s 中 找到所有 p 的变位词。

　　利用两个HashMap 和 Sliding window：

　　　　先把 p 的char 和 出现次数 存入 mapP；

　　　　然后遍历string s，利用 sliding window 把 和 p 一样长度的 string 的 char 保存在 tempMap 里，比较 tempMap 和 mapP。

Java Solution:

Runtime beats 20.00% 

完成日期：11/07/2017

关键词：HashMap

关键点：利用sliding window 把 tempMap 和 mapP 比较

class Solution 
{
    public List<Integer> findAnagrams(String s, String p) 
    {
        List<Integer> list = new ArrayList<>();
        HashMap<Character, Integer> mapP = new HashMap<>();
        HashMap<Character, Integer> tempMap = new HashMap<>();
        
        for(char c: p.toCharArray()) // store p char and occurrence into mapP
            mapP.put(c, mapP.getOrDefault(c, 0) + 1);
        
        
        for(int i=0; i<s.length(); i++) // iterate s and update a tempMap with p len
        {
            char c = s.charAt(i);
            char leftC;
            
            tempMap.put(c, tempMap.getOrDefault(c, 0) + 1);
            
            if(i >= p.length()) // once reach to p's length, remove most left char
            {
                leftC = s.charAt(i - p.length());
                // remove left char
                if(tempMap.get(leftC) == 1)
                    tempMap.remove(leftC);
                else
                    tempMap.put(leftC, tempMap.get(leftC) - 1);
            }
            
            if(tempMap.equals(mapP))
                list.add(i + 1 - p.length());
         
            
        }
            
        return list;
    }
}

参考资料：N/A

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