150.Find All Anagrams in a String（寻找字符串中所有变位词）

题目：

Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.

给定一个字符串s和一个非空字符串p，找到s中p的变位词的所有起始索引。

Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.

字符串仅由小写英文字母组成，字符串s和p的长度不会大于20,100。

The order of output does not matter.

输出顺序无关紧要。

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".

解答：

 自己理解写的：

class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        List<Integer> res=new LinkedList<>();
        if (s == null || s.length() == 0 || p == null || p.length() == 0) return res;
        int left=0,right=0,count=p.length();
        int[] hash=new int[256];
        for(char c:p.toCharArray())
            hash[c]++;
        while(right<s.length()){
            //1
            if(hash[s.charAt(right)]>0) count--;
            hash[s.charAt(right)]--;
            right++;
            //2
            if(count==0) res.add(left);
            //3
            if(right-left==p.length()){
                if(hash[s.charAt(left)]>=0) count++;
                hash[s.charAt(left)]++;
                left++;
            }
        }
        return res;
    }
}

精简写法：

class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        List<Integer> res=new LinkedList<>();
        if (s == null || s.length() == 0 || p == null || p.length() == 0) return res;
        int left=0,right=0,count=p.length();
        int[] hash=new int[256];
        for(char c:p.toCharArray())
            hash[c]++;
        while(right<s.length()){
            if(hash[s.charAt(right++)]-->0) count--;
            if(count==0) res.add(left);
            if(right-left==p.length() && hash[s.charAt(left++)]++>=0) count++;
        }
        return res;
    }
}

详解：

滑动窗口--双指针+hashmap（这里可以用大小为256的数组代替，位置代表该字符ASCII码对应的数字，数组的内容是出现的次数）。

left，right分别表示滑动窗口的左右边界，count表示p中需要匹配的字符个数（也是差异度）。

如果右边界所指的字符已经在hash表中，说明该字符在p中，count减1，hash表中该字符减1，右边界加1。

如果count==0，说明p中的字符都在s中，将left加入结果中。

如果right-left=p的长度，则应将最左边的字符去掉，如果该字符在hashmap中的个数大于等于0，说明该字符在p中，否则原来在hash表中为0，减1后应为-1，如果去掉了属于p的字符，count（差异度）加1。