LeetCode 438. Find All Anagrams in a String

原题链接在这里：https://leetcode.com/problems/find-all-anagrams-in-a-string/

题目：

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". 

题解：

先将p的所有char对应的map count++.

然后快慢指针维护一个sliding window. 右侧runner每次都移动，对应的char的count减1. 若是count减1之前大于0, 表示char在p中, sum--.

sum为0时就找到了anagram.

当window大小为p长度时，移动左侧walker, 对应char的count加1. 若是count加1之前不是负数，表示这个char在p中, sum++.

Time Complexity: O(n). n = s.length().

Space: O(1). 用到了map.

AC Java:

public class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        List<Integer> res = new ArrayList<Integer>();
        if(s == null || p == null || s.length() == 0 || p.length() == 0){
            return res;
        }
        int [] map = new int[256];
        for(int i = 0; i<p.length(); i++){
            map[p.charAt(i)]++;
        }
        int walker = 0;
        int runner = 0;
        int sum = p.length();
        while(runner < s.length()){
            if(map[s.charAt(runner++)]-- > 0){
                //Move runner everytime. Decrease corresponding char count by 1
                //If current char count is larger than 0, then this char is in p
                sum--;
            }
            
            if(sum == 0) {
                //Find anagram
                res.add(walker);
            }

            if(runner-walker == p.length() && map[s.charAt(walker++)]++ >= 0){
                //If windows's size is already p.length(), move walker and increase corresponding char count by 1 
                //If count before increasing is not smaller than 0, this char is in p
                sum++;
            }
        }
        return res;
    }
}

类似Permutation in String.