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# 1277.Count-Square-Submatrices-with-All-Ones

## 题目描述

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.
Constraints:
1 <= arr.length <= 300
1 <= arr.length <= 300
0 <= arr[i][j] <= 1

## 代码

### Approach #1 DP

Time: O(N^2) && Space: O(N^2)
class Solution {
public int countSquares(int[][] matrix) {
int n = matrix.length;
int m = matrix.length;
int[][] dp = new int[n][m];
int count = 0;
for (int i = 0; i < n; i++)
if (matrix[i] == 1)
dp[i] = 1;
for (int j = 0; j < m; j++)
if (matrix[j] == 1)
dp[j] = 1;
for (int i = 1; i < n; i++) {
for (int j = 1; j < m; j++) {
if (matrix[i][j] == 1) {
int min = 1 + Math.min(dp[i - 1][j],
Math.min(dp[i][j - 1], dp[i - 1][j - 1]));
dp[i][j] = min;
}
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
count += dp[i][j];
}
}
return count;
}
}