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  • HackerRank

    It requires O(nlgn) solution. And actually I think the passing code is for "non-decreasing"..

    #include <cmath>
    #include <cstdio>
    #include <vector>
    #include <unordered_set>
    #include <string>
    #include <iostream>
    #include <algorithm>
    using namespace std;
    
    int main() {
        int t; cin >> t;
        vector<int> arr(t);
        for (int i = 0; i < t; i++)
            cin >> arr[i];
    
        vector<int> dp(t, 1);
    
        //    c[i]: Smallest LAST elem of a LIS seq with lenghth i
        vector<int> c(1, arr[0]);
    
        int ret = 1;
        for (int i = 1; i < t; i++)
        {
            if (arr[i] <= c[0])
            {
                c[0] = arr[i];
                dp[i] = 1;
            }
            else if (arr[i] >= c.back())
            {
                c.push_back(arr[i]);
                dp[i] = c.size();
            }
            else
            {
                int k = std::lower_bound(c.begin(), c.end(), arr[i]) - c.begin();
                c[k] = arr[i];
                dp[i] = k + 1;
            }
            ret = std::max(ret, dp[i]);
        }
        cout << ret << endl;
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/tonix/p/4492072.html
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