【2022-06-11-第80场双周赛】

总结

水题，Q1WA两次是逆天。

Q1.强密码检验器 II

模拟即可。

class Solution {
public:
    bool strongPasswordCheckerII(string p) {
        int h[4] = {0};
        string t = "!@#$%^&*()-+";
        if(p.size() < 8) return false;
        for(int i = 0; i < p.size() - 1; ++i){
            if(p[i] == p[i+1]) return false;
        }
        for(auto i : p){
            if(i >= 'a' && i <= 'z') h[1] = 1;
            if(i >= 'A' && i <= 'Z') h[0] = 1;
            if(i >= '0' && i <= '9') h[3] = 1;
            if(t.find(i) != -1) h[2] = 1;
        }
        for(int i = 0; i < 4; ++i) if(h[i] == 0) return false;
        return true;
    }
};

Q2. 咒语和药水的成功对数

class Solution {
public:
    vector<int> successfulPairs(vector<int>& s, vector<int>& p, long long sc) {
        sort(p.begin(), p.end());
        int n = s.size(), m = p.size();
        vector<int> ret;
        for(auto i : s){
            int l = -1, r = m;
            while(l + 1 < r){
                int mid = l + r >> 1;
                if((long long)p[mid] * i < sc) l = mid;
                else r = mid;
            }
            ret.push_back(m - r);
        }
        return ret;
    }
};

Q3.替换字符后匹配

用哈希表记录转换路径，然后暴力匹配。

class Solution {
public:
    bool matchReplacement(string s, string sub, vector<vector<char>>& mps) {
        map<char, set<char>> mp;
        int n = s.size(), m = sub.size();
        int h[n - m + 1]; memset(h, 0, sizeof(h));
        int p[128][128]; memset(p, 0, sizeof(p));
        for(auto &i : mps) p[i[0]][i[1]] = 1;
        for(int i = 0; i < m; ++i){
            for(int j = 0; j <= n - m; ++j){
                if(sub[i] == s[i + j] || p[sub[i]][s[i + j]]){
                    h[j]++;
                }
            }
        }
        for(int i = 0; i <= n - m; ++i){
            if(h[i] == m) return true;
        }
        return false;
    }
};

Q4.https://leetcode.cn/problems/count-subarrays-with-score-less-than-k/

可二分可滑窗。

class Solution {
public:
    long long countSubarrays(vector<int>& a, long long k) {
        int n = a.size();
        long long ps[n + 1]; ps[0] = 0;
        long long ret = 0;
        for(int i = 0; i <n; ++i) ps[i + 1] = ps[i] + a[i];
        for(int i = 1; i <= n; ++i){
            int l = -1, r = i + 1;
            while(l + 1 < r){
                int mid = l + r >> 1;
                if((ps[i] - ps[mid]) * (i - mid) >= k) l = mid;
                else r = mid;
            }
            ret += i - r;
        }
        return ret;
    }
};