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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; }
HackerRank - "The Longest Increasing Subsequence"
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原文地址:http://www.cnblogs.com/tonix/p/4492072.html