随笔分类 -  单调栈/单调队列

摘要:link Solution1: class Solution { public: int m; int n; int numSubmat(vector<vector<int>>& mat) { m=mat.size(); n=mat[0].size(); int res=0; for(int up= 阅读全文
posted @ 2020-07-07 09:39 feibilun 阅读(137) 评论(0) 推荐(0)
摘要:link class Solution { public: int findMaxValueOfEquation(vector<vector<int>>& points, int k) { deque<int> dq; int n=points.size(); int res=INT_MIN; fo 阅读全文
posted @ 2020-06-28 17:15 feibilun 阅读(176) 评论(0) 推荐(0)
摘要:link 题解: 参考 https://www.luogu.com.cn/blog/12cow/SBCOI2020 从i到i-1,dp[k+1][j]与dp[i][k]交点左移,deque后端pop,需要多考虑i-1这个点,比较后加到deque前端。 #include <bits/stdc++.h> 阅读全文
posted @ 2020-05-19 08:17 feibilun 阅读(143) 评论(0) 推荐(0)
摘要:link class Solution { public: int longestSubarray(vector<int>& nums, int limit) { int n=nums.size(); deque<int> maxdq; deque<int> mindq; int right=0; 阅读全文
posted @ 2020-05-03 12:49 feibilun 阅读(192) 评论(0) 推荐(0)
摘要:link 1.dfs+memo: class Solution { public: int n; int maxJumps(vector<int>& arr, int d) { n=arr.size(); vector<int> memo(n,-1); int res=0; for(int i=0; 阅读全文
posted @ 2020-04-27 16:48 feibilun 阅读(125) 评论(0) 推荐(0)
摘要:link using deque: class Solution { public: int constrainedSubsetSum(vector<int>& nums, int k) { int n=nums.size(); int res=nums[0]; deque<int> dq; for 阅读全文
posted @ 2020-04-26 17:03 feibilun 阅读(171) 评论(0) 推荐(0)