力扣 第 314 场周赛

T1 处理用时最长的那个任务的员工

class Solution {
public:
    int hardestWorker(int n, vector<vector<int>>& logs) {
        int pre = 0, maxv = 0;
        int res = 999999;
        for(int i = 0; i < logs.size(); i ++)
        {
            int id = logs[i][0], time = logs[i][1];
            
            if(maxv <= time - pre)
            {
                if(maxv == time - pre) res = min(res, id);
                else res = id;
                maxv = time - pre;
            }
            pre = time;
        }
        return res;
    }
};
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T2 找出前缀异或的原始数组

class Solution {
public:
    vector<int> findArray(vector<int>& a) {
        vector<int> res;
        int t = 0, sum = 0;
        for(auto x : a)
        {
            t = x ^ sum;
            res.push_back(t);
            sum ^= t;
        }
        return res;
    }
};
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T3 使用机器人打印字典序最小的字符串

class Solution {
public:
    int cnt[30] = {0};
    
    bool check(char c)
    {
        int t = c - 'a';
        for(int i = 0; i < t; i ++)
        {
            if(cnt[i]) return false;
        }
        return true;
    }
    
    string robotWithString(string s) {
        string res = "";
        
        for(int i = 0; i < s.size(); i ++)
            cnt[s[i] - 'a'] ++;
        
        stack<char> stk;
        stk.push(s[0]);
        cnt[s[0] - 'a'] --;
        
        for(int i = 1; i < s.size(); i ++)
        {
            while(stk.size())
            {
                cnt[stk.top() - 'a'] ++;
                if(check(stk.top()))
                {
                    res += stk.top();
                    cnt[stk.top() - 'a'] --;
                    stk.pop();
                }
                else 
                {
                    cnt[stk.top() - 'a'] --;
                    break;
                }
            }
            stk.push(s[i]);
            cnt[s[i] - 'a'] --;
        }
        
        while(stk.size()) res += stk.top(), stk.pop();
        return res;
    }
};
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T4 矩阵中和能被 K 整除的路径

class Solution {
public:
    int mod = 1e9 + 7;
    int numberOfPaths(vector<vector<int>>& a, int k) {
        int n = a.size(), m = a[0].size();
        vector<vector<int>> g (n+1,vector<int>(m + 1, 0));
        
        for(int i = 0; i < n; i ++)
            for(int j = 0; j < m; j ++)
                g[i + 1][j + 1] = a[i][j];
        
        vector<vector<vector<int>>> f(n+1,vector<vector<int>>(m+1,vector<int>(k + 1, 0)));
        
        f[1][1][g[1][1] % k] = 1;
        
        for(int i = 1; i <= n; i ++)
            for(int j = 1; j <= m; j ++)
                for(int u = 0; u < k; u ++)
                    if(i == 1 && j == 1) continue;
                    else f[i][j][(u + g[i][j]) % k] = (f[i - 1][j][u] + f[i][j - 1][u]) % mod;
        
        return f[n][m][0];
    }
};
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