***Leetcode 354. Russian Doll Envelopes

https://leetcode.com/problems/russian-doll-envelopes/description/

LIS就能过掉

然后暴力O(n^2)的写法

bool cmp_small(pair a, pairb) {
    if (a.first == b.first) return a.second > b.second;
    else return a.first < b.first;
}

class Solution {
public:
    int maxEnvelopes(vector>& envelopes) {
        if (envelopes.size() == 1) return 1;
        sort(envelopes.begin(), envelopes.end(), cmp_small);
        vector dp(envelopes.size()+1, 0);
        int ans = 0;
        for (int i = 0; i < envelopes.size(); i++) {
            bool ch = false;
            for (int j = 0; j < i; j++) {
                if ( envelopes[j].first < envelopes[i].first && envelopes[j].second < envelopes[i].second )
                        dp[i] = max(dp[i], dp[j] + 1), ch = true;
                 
                
            }
            if (!ch) dp[i] = 1;
            ans = max(ans, dp[i]);
        }
        return ans;
    }
};

另外LIS有O(nlog(n))的写法：

LIS的nlogn的方程 dp[i]表示，最长公共子序列 第[i]个的最小值

bool cmp_small(pair a, pairb) {
    if (a.first == b.first) return a.second > b.second;
    else return a.first < b.first;
}

class Solution {
public:
    int maxEnvelopes(vector< pair >& ens) {
        if(ens.size() <= 1) return ens.size();
        sort(ens.begin(), ens.end(), cmp_small);
        // int dp[ens.size()+1] = {0};
        vector< int > dp;
        for (auto v : ens) {
            int pos = lower_bound( dp.begin(), dp.end(), v.second ) - dp.begin() ;
            if (pos >= dp.size()) {
                dp.push_back(v.second);
            } else {
                dp[pos]  = v.second;
            }
            
        }
        return dp.size();
    }
};