[leetcode]438. Find All Anagrams in a String找出所有变位词

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.

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".

 

思路：

sliding window滑动窗口

 

1.  traverse String T,  using a map to record each char's frequency

2.  use [leftMost to i] to maintain a sliding window, making sure that each char's frequency in such sliding window == that in T

3.  if mapS [S.charAt(leftMost)]  >  mapT [S.charAt(leftMost)] ,  it signs we can move sliding window

4.  how to find the next sliding window?  move leftMost, meanwhile,  decrement mapS [S.charAt(leftMost)]  until we find each frequency in  [leftMost to i] == that in T 

 

 

代码：

class Solution {
    public List<Integer> findAnagrams(String S, String T) {
        List<Integer> resultList = new ArrayList<>();
        if (S == null || S.length() == 0) return resultList;

        int[] mapT = new int[256];
        int[] mapS = new int[256];

        int count = 0;
        int leftMost = 0;
        for(int i = 0; i < T.length(); i++){
            mapT[T.charAt(i)] ++;
        }

        for(int i = 0; i < S.length(); i++) {
            char s = S.charAt(i);
            mapS[s]++;
            if (mapT[s] >= mapS[s]) {
                count++;
            }

            if (count == T.length()) {
                while (mapS[S.charAt(leftMost)] > mapT[S.charAt(leftMost)]) {
                    if (mapS[S.charAt(leftMost)] > mapT[S.charAt(leftMost)]) {
                        mapS[S.charAt(leftMost)]--;
                    }
                    leftMost++;
                }
                if (i - leftMost + 1 == T.length()) {
                    resultList.add(leftMost);
                }
            }
        }
        return resultList;
    }
}