力扣438. 找到字符串中所有字母异位词（滑动窗口）

 常规的滑动窗口问题，要注意满足答案的条件：1.窗口长度与p字符串长度一致2.字符充足的个数与p中不重复个数相同

class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        vector<int> res;
        unordered_map<char, int> window, need;
        for (auto i : p) {
            need[i]++;
        }
        int left = 0, right = 0;
        int enough = 0; //记录窗口中字符数量充足的个数，注意：字符充足并不完全符合题意，需要字符数量完全一致才行
        while (right < s.length()) {
            if (need.count(s[right])) {
                window[s[right]]++;
                if (window[s[right]] == need[s[right]]) {
                    enough++;
                }
            }
            right++;
            while (enough == need.size()) {
                if (need.count(s[left])) {
                    if (right - left == p.size()) { //窗口长度等于p的长度
                        res.push_back(left);
                    }
                    if (window[s[left]] == need[s[left]]) {
                        enough--;
                    }
                    window[s[left]]--;
                }
                left++;
            }
        }
        return res;
    }
};