力扣HOT100-找到字符串中所有字母异位词

思路

维护一个长度与p一致的窗口，记录即可。

Python版本

class Solution:
    def findAnagrams(self, s: str, p: str) -> List[int]:
        n,m = len(s),len(p)
        if n < m:
            return []
        s_record = [0] * 26
        p_record = [0] * 26
        res = []
        # 记录目标集合
        for c in p:
             p_record[ord(c) - 97] += 1
        # 构建一个窗口
        left,right = 0,0
        for i in range(m):
            s_record[ord(s[right]) - 97] += 1
            right += 1
        if s_record == p_record:
            res.append(left)
        # 移动窗口
        while right < n:
            s_record[ord(s[left]) - 97] -= 1
            left += 1
            s_record[ord(s[right]) - 97] += 1
            right += 1
            if s_record == p_record:
                res.append(left)
        return res
        

Java版本

class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        int n = s.length();
        int m = p.length();
        List<Integer> res = new ArrayList<Integer>();
        int[] p_record = new int[26];
        int[] s_record = new int[26];
        if (m > n){
            return res;
        }
        // 存储p的元素
        for (int i = 0;i<m;i++){
            p_record[p.charAt(i) - 'a']++;
        }
        // 构造窗口
        int left = 0;
        int right;
        for (right = 0;right<m;right++){
            s_record[s.charAt(right) - 'a']++;
        }
        if (Arrays.equals(s_record, p_record)){
            res.add(left);
        }
        while (right < n){
            s_record[s.charAt(left) - 'a']--;
            left++;
            s_record[s.charAt(right) - 'a']++;
            right++;
            if (Arrays.equals(s_record, p_record)){
            res.add(left);
            }
        }
        return res;
    }
}