[leetcode] 1504. Count Submatrices With All Ones

原创

是念 2022-08-11 17:13:32 博主文章分类：leetcode题解 ©著作权

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Description

Given a rows * columns matrix mat of ones and zeros, return how many submatrices have all ones.

Example 1:

Input: mat = [[1,0,1],
              [1,1,0],
              [1,1,0]]
Output: 13
Explanation:
There are 6 rectangles of side 1x1.
There are 2 rectangles of side 1x2.
There are 3 rectangles of side 2x1.
There is 1 rectangle of side 2x2. 
There is 1 rectangle of side 3x1.
Total number of rectangles = 6 + 2 + 3 + 1 + 1 = 13.

Example 2:

Input: mat = [[0,1,1,0],
              [0,1,1,1],
              [1,1,1,0]]
Output: 24
Explanation:
There are 8 rectangles of side 1x1.
There are 5 rectangles of side 1x2.
There are 2 rectangles of side 1x3. 
There are 4 rectangles of side 2x1.
There are 2 rectangles of side 2x2. 
There are 2 rectangles of side 3x1. 
There is 1 rectangle of side 3x2. 
Total number of rectangles = 8 + 5 + 2 + 4 + 2 + 2 + 1 = 24.

Example 3:

Input: mat = [[1,1,1,1,1,1]]
Output: 21

Example 4:

Input: mat = [[1,0,1],[0,1,0],[1,0,1]]
Output: 5

Constraints:

1 <= rows <= 150
1 <= columns <= 150
0 <= mat[i][j] <= 1

分析

题目的意思是：这道题要求出1组成的小矩形，这道题暴力求解肯定会完蛋，我参考了一下别人的思路，首先rows[i][j]表示第i行以j结尾的连续1的个数。然后再遍历该矩阵，统计以i，j为右下角的矩阵的个数，由于我们能够通过rows数组知道以i,j为结尾的连续1的个数，k从第i行到第0行，一个一个的找符合条件的就行了,其中rows[k][j]和rows[i][j]围成的矩形的个数位min(rows[k][j],rows[i][j]),不清楚的话，自己算一个哈哈哈。

代码

class Solution:
    def numSubmat(self, mat: List[List[int]]) -> int:
        m=len(mat)
        n=len(mat[0])
        rows=[[0]*n for i in range(m)]
        for i in range(m):
            for j in range(n):
                if(j==0):
                    rows[i][j]=mat[i][j]
                else:
                    if(mat[i][j]==0):
                        rows[i][j]=0
                    else:
                        rows[i][j]=rows[i][j-1]+1
        res=0
        for i in range(m):
            for j in range(n):
                min_num=rows[i][j]
                for k in range(i,-1,-1):
                    min_num=min(min_num,rows[k][j])
                    if(min_num==0):
                        break
                    res+=min_num
        return res