LeetCode第332场周赛

2023.2.12LeetCode第332场周赛

2562. 找出数组的串联值

思路

双指针模拟，两个指针相遇的时候要特判

算法

class Solution {
public:
    long long findTheArrayConcVal(vector<int>& nums) {
        long long ans = 0;
        int i = 0, j = nums.size() - 1;
        while (i <= j) {
            if (i != j) {
                string s = to_string(nums[i]) + to_string(nums[j]);
                ans += stoll(s);
            } else {
                string s = to_string(nums[i]);
                ans += stoll(s);
                break;
            }
            i ++ , j -- ;
        }
        return ans;
    }
};

2563. 统计公平数对的数目

思路

计算在[lower, upper]之间满足条件的数目可以利用前缀和的思想转化为不超过upper和不超过lower-1的数量相减
遍历所有nums[j]，找到前面的满足条件的nums[i]的数量，需要先对数组排序后进行二分

算法

class Solution {
public:
    typedef long long ll;
    
    ll calc(vector<int> &a, ll x) {
        ll res = 0;
        int n = a.size();
        for (int i = 1; i < a.size(); i ++ ) {
            int l = 0, r = i - 1;
            while (l < r) {
                int mid = l + r + 1 >> 1;
                if (a[i] + a[mid] <= x) l = mid;
                else r = mid - 1;
            }
            if (a[i] + a[l] <= x)
                res += l + 1;
        }
        return res;
    }
    
    long long countFairPairs(vector<int>& a, int lower, int upper) {
        sort(a.begin(), a.end());
        return calc(a, upper) - calc(a, lower - 1);
    }
};

2564. 子字符串异或查询

思路

由于val ^ first == second，根据异或运算性质，val = first ^ second
即找到val的二进制在给定字符串中最先出现的位置
暴力算法是记录s的所有子串，o(n2)会超时
由于二进制长度不会超过32位，故记录时只用对每个起点枚举32位的长度
最后查找结果即可

算法

class Solution {
public:
    vector<vector<int>> substringXorQueries(string s, vector<vector<int>>& queries) {
        vector<vector<int>> ans;
        unordered_map<string, int> mp;
        int n = s.size();
        for (int i = 0; i < n; i ++ )
            for (int j = i; j < n; j ++ ) {
                string t = s.substr(i, j - i + 1);
                if (j - i + 1 >= 32) break;
                if (!mp.count(t))
                    mp[t] = i;
            }
        for (auto x : queries) {
            int t = x[0] ^ x[1];
            string a;
            while (t) {
                if (t & 1) a = '1' + a;
                else a = '0' + a;
                t >>= 1;
            }
            if (a.size() == 0) a = "0";
            if (mp.count(a)) ans.push_back({mp[a], mp[a] + (int)a.size() - 1});
            else ans.push_back({-1, -1});
        }
        return ans;
    }
};