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English Version

题目描述

给你一个 m * n 的矩阵,矩阵中的元素不是 0 就是 1,请你统计并返回其中完全由 1 组成的 正方形 子矩阵的个数。

 

示例 1:

输入:matrix =
[
  [0,1,1,1],
  [1,1,1,1],
  [0,1,1,1]
]
输出:15
解释: 
边长为 1 的正方形有 10 个。
边长为 2 的正方形有 4 个。
边长为 3 的正方形有 1 个。
正方形的总数 = 10 + 4 + 1 = 15.

示例 2:

输入:matrix = 
[
  [1,0,1],
  [1,1,0],
  [1,1,0]
]
输出:7
解释:
边长为 1 的正方形有 6 个。 
边长为 2 的正方形有 1 个。
正方形的总数 = 6 + 1 = 7.

 

提示:

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

解法

Python3

class Solution:
    def countSquares(self, matrix: List[List[int]]) -> int:
        m, n = len(matrix), len(matrix[0])
        f = [[0] * n for _ in range(m)]
        ans = 0
        for i, row in enumerate(matrix):
            for j, v in enumerate(row):
                if v == 0:
                    continue
                if i == 0 or j == 0:
                    f[i][j] = 1
                else:
                    f[i][j] = min(f[i - 1][j - 1], f[i - 1][j], f[i][j - 1]) + 1
                ans += f[i][j]
        return ans

Java

class Solution {
    public int countSquares(int[][] matrix) {
        int m = matrix.length;
        int n = matrix[0].length;
        int[][] f = new int[m][n];
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (matrix[i][j] == 0) {
                    continue;
                }
                if (i == 0 || j == 0) {
                    f[i][j] = 1;
                } else {
                    f[i][j] = Math.min(f[i - 1][j - 1], Math.min(f[i - 1][j], f[i][j - 1])) + 1;
                }
                ans += f[i][j];
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int countSquares(vector<vector<int>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        int ans = 0;
        vector<vector<int>> f(m, vector<int>(n));
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (matrix[i][j] == 0) continue;
                if (i == 0 || j == 0)
                    f[i][j] = 1;
                else
                    f[i][j] = min(f[i - 1][j - 1], min(f[i - 1][j], f[i][j - 1])) + 1;
                ans += f[i][j];
            }
        }
        return ans;
    }
};

Go

func countSquares(matrix [][]int) int {
	m, n, ans := len(matrix), len(matrix[0]), 0
	f := make([][]int, m)
	for i := range f {
		f[i] = make([]int, n)
	}
	for i, row := range matrix {
		for j, v := range row {
			if v == 0 {
				continue
			}
			if i == 0 || j == 0 {
				f[i][j] = 1
			} else {
				f[i][j] = min(f[i-1][j-1], min(f[i-1][j], f[i][j-1])) + 1
			}
			ans += f[i][j]
		}
	}
	return ans
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

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