438. Find All Anagrams in a String

Posted 2020-10-11 andywu

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".
题目含义：给出一个字符串s和一个非空的字符串p，找到p的重组字在s中出现的开始位置

 1     public List<Integer> findAnagrams(String s, String p) {
 2 //        http://blog.csdn.net/chenwuji91/article/details/52981530
 3         List<Integer> result = new ArrayList<>();
 4         if (s == null || s.length() == 0 || p == null || p.length() == 0)
 5             return result;
 6         int[] hash = new int[256];
 7         char[] pp = p.toCharArray();
 8         for (char i : pp) {
 9             hash[i]++;
10         }
11         int left = 0, right = 0, count = p.length();
12         while (right < s.length()) {
13             if (hash[s.charAt(right++)]-- > 0)  //窗口右移；相应的hash值减小；如果这个位置的Hash值是正的，表示p字符串也包含这个，所以count做减法
14                 count--;
15             if (count == 0)
16                 result.add(left);//count指示器，为0表示和p对应的hash值完全一致
17             if (right - left == p.length() && hash[s.charAt(left++)]++ >= 0)
18                 //如果当窗口大小和需要比较的字符串大小一致的时候，将窗口左边的指针向右边移动，移动的同时左边的字符计数因为在第一个if的地方hash值减小过，所以需要执行对应恢复操作，即：hash值增加，count计数值增加。
19                 count++;
20         }
21         return result;        
22     }