JavaC++算法题解leetcode1582二进制矩阵特殊位置

2024-04-02 19:04:59 758人浏览安东尼

摘要

目录题目要求思路：模拟Javac++Rust题目要求思路：模拟直接按题意模拟，先算出每行每列中“111”的个数，然后判断统计行列值均为111的位置即可

题目要求

思路：模拟

直接按题意模拟，先算出每行每列中“111”的个数，然后判断统计行列值均为111的位置即可。

Java

class Solution {
    public int numSpecial(int[][] mat) {
        int n = mat.length, m = mat[0].length;
        int res = 0;
        int[] row = new int[n], col = new int[m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                row[i] += mat[i][j];
                col[j] += mat[i][j];
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (mat[i][j] == 1 && row[i] == 1 && col[j] == 1)
                    res++;
            }
        }
        return res;
    }
}

时间复杂度：O(m×n)
空间复杂度：O(m+n)

C++

class Solution {
public:
    int numSpecial(vector<vector<int>>& mat) {
        int n = mat.size(), m = mat[0].size();
        int res = 0;
        vector<int> row(n), col(m);
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                row[i] += mat[i][j];
                col[j] += mat[i][j];
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (mat[i][j] == 1 && row[i] == 1 && col[j] == 1)
                    res++;
            }
        }
        return res;
    }
};

时间复杂度：O(m×n)
空间复杂度：O(m+n)

Rust

这里的迭代函数用得不是很熟练，参考了好多才勉强理解下来。

impl Solution {
    pub fn num_special(mat: Vec<Vec<i32>>) -> i32 {
        let row = mat.iter().map(|row| row.iter().sum::<i32>()).collect::<Vec<_>>();
        let col = (0..mat[0].len()).map(|i| mat.iter().map(|col| col[i]).sum::<i32>()).collect::<Vec<_>>();
        (0..mat.len()).fold(0, |res, i| res + (0..mat[i].len()).filter(|&j| mat[i][j] == 1 && row[i] == 1 &&col[j] == 1).count() as i32)
    }
}