描述
Given a m * n matrix of ones and zeros, return how many square submatrices have all ones.
Example 1:
Input: matrix =
[
[0,1,1,1],
[1,1,1,1],
[0,1,1,1]
]
Output: 15
Explanation:
There are 10 squares of side 1.
There are 4 squares of side 2.
There is 1 square of side 3.
Total number of squares = 10 + 4 + 1 = 15.
Example 2:
Input: matrix =
[
[1,0,1],
[1,1,0],
[1,1,0]
]
Output: 7
Explanation:
There are 6 squares of side 1.
There is 1 square of side 2.
Total number of squares = 6 + 1 = 7.
Note:
1 <= arr.length <= 300
1 <= arr[0].length <= 300
0 <= arr[i][j] <= 1
解析
根据题意，只需要找出 matrix 中由 1 组成的任意大小的正方形个数即可。其实只要找规律就可以发现，设计一个 m*n 的 new 矩阵来存放数量值，其中的 new[i][j] 表示以此单元格为右下角的正方形的个数可以有几个，然后遍历比较所有的 matrix[1:,1:] 的值，如果 matrix[i][j] 值为 1 ，则说明至少有一个边长为 1 的正方形，然后比较 grid[i-1][j-1]、grid[i][j-1]、grid[i-1][j] 三个中的最小值 min ，表示可以至少在以 [i,j] 为右下角的地方可以形成 min 个正方形，将 min+1 赋值给 gridp[i][j] 即可，如果完成将值填充到 grid 对应的位置上面，最后将所有的 new 元素求和即可，
解答
class Solution(object):
def countSquares(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: int
"""
import numpy as np
M = len(matrix)
N = len(matrix[0])
grid = [[0] * N for _ in range(M)]
for i in range(N):
grid[0][i] = matrix[0][i]
for j in range(M):
grid[j][0] = matrix[j][0]
for i in range(1, M):
for j in range(1, N):
if matrix[i][j] == 1:
grid[i][j] = min(grid[i-1][j-1] , grid[i-1][j] , grid[i][j-1]) + 1
return np.sum(grid)
运行结果
Runtime: 520 ms, faster than 83.00% of Python online submissions for Count Square Submatrices with All Ones.
Memory Usage: 15.2 MB, less than 81.30% of Python online submissions for Count Square Submatrices with All Ones.
解析
还可以进行三重循环遍历来找可能合法的正方形，但是这种方法肯定超时。
解答
class Solution(object):
def countSquares(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: int
"""
import numpy as np
M = len(matrix)
N = len(matrix[0])
L = min(M,N)
matrix = np.array(matrix)
result = 0
for l in range(1, L+1):
for i in range(M+1-l):
for j in range(N+1-l):
if np.sum(matrix[i:i+l,j:j+l]) == l**2:
result += 1
return result
运行结果
Time Limit Exceeded
原题链接：leetcode.com/problems/co…
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