567. Permutation in String

滑動窗口的題目, 我們先把s1的所有字母數量存起來

然後在滑動窗口的左邊條件是當r, l的距離已經大於s1的size就該移動了,

因為這個窗口的size剛好就是s1的size

class Solution {
public:
  bool checkInclusion(string s1, string s2) {
    vector<int> count2(26, 0);
    vector<int> count1(26, 0);
    for(auto c: s1){
      count1[c - 'a']++;
    }
    int l = 0;
    int r = 0;
    while(r < s2.size()){
      char c = s2[r];
      count2[c - 'a']++;
      if(count1 == count2){
        return true;
      }
      r++;
      while(r - l >= s1.size()){
        count2[s2[l] - 'a']--;
        l++;
      }
    }
    return false;
  }
};