力扣1504.统计全1子矩形 开一个二维数组存每个点从它本身开始向左有多少连续的1 遍历矩形右下角(i,j) 再遍历行k in i每一行的矩形数量 minx = min(minx,left(k,j)) class Solution {public:int numSubmat(vector<vector<int>>& mat) {int n = mat.size();int m = mat
1 TOP Eleven 1504 Tips 1.1 The questions you ask 1 The questions you ask will very often determine the quest that you take. 2 And if we only ask the negative questions such as “Why d
统计全1子矩阵个数 思路1:首先考虑深度优先模拟,从【0,0】出发向下、右扩展,符合条件res++,最后输出res,比较直观,但重复进行了大量节点遍历操作,时间复杂度较高,数据量大时会超时 class Solution {unordered_set<int>set;int res=0;void get(vector<vector<int>>& mat,int start_r,int star
哈佛幸福课11条总结 Top Eleven 1504 Tips Oh, the questions you’ll ask Questions will very often determine the quest that you take We also need to ask the positive question Questions start quest Believ