LeetCode每日一题(Count Square Submatrices with All Ones)

最新推荐文章于 2023-01-23 18:05:40 发布

原创最新推荐文章于 2023-01-23 18:05:40 发布 · 103 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #动态规划 #算法

算法专栏收录该内容

413 篇文章

订阅专栏

本文介绍了一种算法，用于计算给定的二进制矩阵中全为1的子矩阵数量，通过两个示例详细展示了如何使用 Rust 代码实现，并讨论了相关约束。主要涉及动态规划的思想，适用于计算机科学和编程挑战。

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

不知道该怎么讲，大家看代码吧

代码实现(Rust):

impl Solution {
    pub fn count_squares(matrix: Vec<Vec<i32>>) -> i32 {
        let mut ans = 0;
        let mut dp = vec![vec![0; matrix[0].len()]; matrix.len()];
        for r in 0..matrix.len() {
            for c in 0..matrix[0].len() {
                if matrix[r][c] == 0 {
                    dp[r][c] = 0;
                } else {
                    if r == 0 || c == 0 {
                        if matrix[r][c] == 1 {
                            ans += 1;
                        }
                        dp[r][c] = matrix[r][c];
                    } else {
                        let count = dp[r-1][c-1].min(dp[r-1][c]).min(dp[r][c-1]) + 1;
                        dp[r][c] = count;
                        ans += count;
                    }
                }
            }
        }
        ans
    }
}