1139.最大的以1为边界的正方形

链接：1139.最大的以1为边界的正方形
难度：Medium
标签：数组、动态规划、矩阵
简介：给你一个由若干 0 和 1 组成的二维网格 grid，请你找出边界全部由 1 组成的最大 正方形 子网格，并返回该子网格中的元素数量。如果不存在，则返回 0。

题解 1 - rust

• 编辑时间：2023-02-17
• 执行用时：4ms
• 内存消耗：2.5MB
• 编程语言：rust
• 解法介绍：同上。

impl Solution {
    pub fn largest1_bordered_square(grid: Vec<Vec<i32>>) -> i32 {
        const MAX: usize = 105;
        let n = grid.len();
        let m = grid[0].len();
        let mut cnt = 0;
        let mut cache = [[[0; 4]; MAX]; MAX];
        for i in 0..n {
            cnt = 0;
            for j in 0..m {
                cache[i][j][0] = cnt;
                cnt = if grid[i][j] == 1 { cnt + 1 } else { 0 };
            }
            cnt = 0;
            for j in (0..m).rev() {
                cache[i][j][1] = cnt;
                cnt = if grid[i][j] == 1 { cnt + 1 } else { 0 };
            }
        }
        for j in 0..m {
            cnt = 0;
            for i in 0..n {
                cache[i][j][2] = cnt;
                cnt = if grid[i][j] == 1 { cnt + 1 } else { 0 };
            }
            cnt = 0;
            for i in (0..n).rev() {
                cache[i][j][3] = cnt;
                cnt = if grid[i][j] == 1 { cnt + 1 } else { 0 };
            }
        }
        cnt = 0;
        for i in 0..n {
            for j in 0..m {
                if grid[i][j] == 0 {
                    continue;
                }
                cnt = cnt.max(1);
                for k in 1..=cache[i][j][1].min(cache[i][j][3]) {
                    if cache[i + k][j][1] >= k && cache[i][j + k][3] >= k {
                        cnt = cnt.max((k + 1).pow(2));
                    }
                }
            }
        }
        cnt as i32
    }
}

题解 2 - python

• 编辑时间：2023-02-17
• 执行用时：272ms
• 内存消耗：16.3MB
• 编程语言：python
• 解法介绍：同上。

class Solution:
    def largest1BorderedSquare(self, grid: List[List[int]]) -> int:
        n, m, cnt = len(grid), len(grid[0]), 0
        MAX = 105
        cache = [[[0] * 4 for _ in range(MAX)] for _ in range(MAX)]
        for i in range(n):
            cnt = 0
            for j in range(m):
                cache[i][j][0] = cnt
                cnt = cnt + 1 if grid[i][j] == 1 else 0
            cnt = 0
            for j in range(m - 1, -1, -1):
                cache[i][j][1] = cnt
                cnt = cnt + 1 if grid[i][j] == 1 else 0
        for j in range(m):
            cnt = 0
            for i in range(n):
                cache[i][j][2] = cnt
                cnt = cnt + 1 if grid[i][j] == 1 else 0
            cnt = 0
            for i in range(n - 1, -1, -1):
                cache[i][j][3] = cnt
                cnt = cnt + 1 if grid[i][j] == 1 else 0
        cnt = 0
        for i in range(n):
            for j in range(m):
                if grid[i][j] == 0:
                    continue
                cnt = max(cnt, 1)
                for k in range(1, min(cache[i][j][1], cache[i][j][3]) + 1):
                    if cache[i + k][j][1] >= k and cache[i][j + k][3] >= k:
                        cnt = max(cnt, pow(k + 1, 2))
        return cnt

题解 3 - cpp

• 编辑时间：2023-02-17
• 执行用时：12ms
• 内存消耗：10.6MB
• 编程语言：cpp
• 解法介绍：预处理每个点的最大延长后遍历。

#define MAX 105
class Solution {
public:
    int largest1BorderedSquare(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size();
        // 0: l, 1: r, 2 : t, 3: b
        int cache[MAX][MAX][4] = {0}, cnt;
        for (int i = 0; i < n; i++) {
            cnt = 0;
            for (int j = 0; j < m; j++) {
                cache[i][j][0] = cnt;
                cnt = grid[i][j] == 1 ? cnt + 1 : 0;
            }
            cnt = 0;
            for (int j = m - 1; j >= 0; j--) {
                cache[i][j][1] = cnt;
                cnt = grid[i][j] == 1 ? cnt + 1 : 0;
            }
        }
        for (int j = 0; j < m; j++) {
            cnt = 0;
            for (int i = 0; i < n; i++) {
                cache[i][j][2] = cnt;
                cnt = grid[i][j] == 1 ? cnt + 1 : 0;
            }
            cnt = 0;
            for (int i = n - 1; i >= 0; i--) {
                cache[i][j][3] = cnt;
                cnt = grid[i][j] == 1 ? cnt + 1 : 0;
            }
        }
        cnt = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 0) continue;
                cnt = max(cnt, 1);
                for (int k = 1; k <= min(cache[i][j][1], cache[i][j][3]); k++) {
                    if (cache[i + k][j][1] >= k && cache[i][j + k][3] >= k) cnt = max(cnt, (int)pow(k + 1, 2));
                }
            }
        }
        return cnt;
    }
};