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[0].length <= 300
0 <= arr[i][j] <= 1
class Solution { public int countSquares(int[][] matrix) { int r = matrix.length; int c = r > 0 ? matrix[0].length : 0; if(r == 0 || c == 0) return 0; int res = 0; int[][] dp = new int[r+1][c+1]; for(int i = 1; i <= r; i++){ for(int j = 1; j <= c; j++){ if(matrix[i-1][j-1] == 1){ dp[i][j] = Math.min(Math.min(dp[i-1][j-1], dp[i][j-1]), dp[i-1][j]) + 1; } if(dp[i][j] != 0) res += dp[i][j]; } } return res; } }
跟https://www.cnblogs.com/wentiliangkaihua/p/11896758.html 221.Maximal Square一毛一样,不过要注意dp数组代表几就要加几,比如3,里面也有1和2维所以要加3.否则会出现考虑情况。