【力扣：1504】统计全1子矩阵

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 start_c,int row,int col){if(row>=mat.size()||col>=mat[0].size()||set.count(start_r+(start_c+((row+col*151)*151))*151)) return;for(int i=start_r;i<=row;i++){if(!mat[i][col]) return;}for(int i=start_c;i<=col;i++){if(!mat[row][i]) return;}res++;set.insert(start_r+(start_c+((row+col*151)*151))*151);get(mat,start_r,start_c,row+1,col);get(mat,start_r,start_c,row,col+1);}
public:int numSubmat(vector<vector<int>>& mat) {for(int i=0;i<mat.size();i++){for(int j=0;j<mat[0].size();j++){get(mat,i,j,i,j);}}return res;}
};

思路2：单考虑行或列时每增加1个1，结果增加 行或列1个数+1，那么多行多列时每增加一行或一列增加（1+2+…+n）*（m+1），加列时：n为行数，m为原来列数，实际上情景就是第一个图的拓展，只不过矩形中的1实际上是长度相等的全1矩形

因而仅需要使用一个二维数组tmp存储target[i][j]及前有几个连续的1，然后从上到下加上min(tmp[i][j],tmp_pre_min)即可

class Solution {
public:int numSubmat(vector<vector<int>>& mat) {int n = mat.size();int m = mat[0].size();vector<vector<int> > row(n, vector<int>(m, 0));for (int i = 0; i < n; ++i) {for (int j = 0; j < m; ++j) {if (j == 0) {row[i][j] = mat[i][j];} else if (mat[i][j]) {row[i][j] = row[i][j - 1] + 1;}else {row[i][j] = 0;}}}int ans = 0;for (int i = 0; i < n; ++i) {for (int j = 0; j < m; ++j) {int col = row[i][j];for (int k = i; k >= 0 && col; --k) {col = min(col, row[k][j]);ans += col;}}}return ans;}
};

单调栈优化后代码：

class Solution {
public:int numSubmat(vector<vector<int>>& mat) {int n = mat.size();int m = mat[0].size();vector<vector<int> > row(n, vector<int>(m, 0));for (int i = 0; i < n; ++i) {for (int j = 0; j < m; ++j) {if (j == 0) {row[i][j] = mat[i][j];} else if (mat[i][j]) {row[i][j] = row[i][j - 1] + 1;}else {row[i][j] = 0;}}}int ans = 0;for (int j = 0; j < m; ++j) { int i = 0; stack<pair<int, int> > Q; int sum = 0; while (i <= n - 1) { int height = 1; while (!Q.empty() && Q.top().first > row[i][j]) {// 弹出的时候要减去多于的答案sum -= Q.top().second * (Q.top().first - row[i][j]); height += Q.top().second; Q.pop(); } sum += row[i][j]; ans += sum; Q.push({ row[i][j], height }); i++; } } return ans;}
};