0073.矩阵置零

https://leetcode.cn/problems/set-matrix-zeroes/

方法一：使用两个标记变量

时间复杂度 $O(m \times n)$，空间复杂度 $O(1)$，m，n 分别表示 matrix 的行数和列数。

func setZeroes(matrix [][]int) {
	m, n, zero_r, zero_c := len(matrix), len(matrix[0]), false, false
	for i := 0; i < m; i++ {
		if matrix[i][0] == 0 {
			zero_r = true
			break
		}
	}
	for i := 0; i < n; i++ {
		if matrix[0][i] == 0 {
			zero_c = true
			break
		}
	}
	for i := 1; i < m; i++ {
		for j := 1; j < n; j++ {
			if matrix[i][j] == 0 {
				matrix[i][0], matrix[0][j] = 0, 0
			}
		}
	}
	for i := 1; i < m; i++ {
		for j := 1; j < n; j++ {
			if matrix[i][0] == 0 || matrix[0][j] == 0 {
				matrix[i][j] = 0
			}
		}
	}
	for i := 0; i < m && zero_r; i++ {
		matrix[i][0] = 0
	}
	for i := 0; i < n && zero_c; i++ {
		matrix[0][i] = 0
	}
}

impl Solution {
    pub fn set_zeroes(matrix: &mut Vec<Vec<i32>>) {
        use std::cmp::{max, min};
        let (m, n, mut zero_r, mut zero_c) = (matrix.len(), matrix[0].len(), false, false);
        for i in 0..max(m, n) {
            if matrix[min(i, m - 1)][0] == 0 {
                zero_r = true;
            }
            if matrix[0][min(i, n - 1)] == 0 {
                zero_c = true;
            }
        }
        for i in 1..m {
            for j in 1..n {
                if matrix[i][j] == 0 {
                    matrix[i][0] = 0;
                    matrix[0][j] = 0;
                }
            }
        }
        for i in 1..m {
            for j in 1..n {
                if matrix[i][0] == 0 || matrix[0][j] == 0 {
                    matrix[i][j] = 0;
                }
            }
        }
        for i in 0..m {
            if zero_r {
                matrix[i][0] = 0;
            }
        }
        for j in 0..n {
            if zero_c {
                matrix[0][j] = 0;
            }
        }
    }
}