73. Set Matrix Zeroes

LeetCode 73. Set Matrix Zeroes

Description

Given an m x n matrix. If an element is 0, set its entire row and column to 0. Do it in-place.

Follow up:

• A straight forward solution using O(mn) space is probably a bad idea.

• A simple improvement uses O(m + n) space, but still not the best solution.

• Could you devise a constant space solution?

Tags

Array

Solution

直接的解法（见解法2）是用两个分别与行和列等长的布尔数组标记对应行列是否存在0，再遍历一遍所有元素，检查其行列号在两数组的对应位置上是否已被标记，如果至少有一个被标记则修改该位置元素为0。

进阶的解法（见解法1）是采用第一行和第一列作为这两个数组，但考虑到二者可能自身就包含0，所以先使用两个布尔变量来记录二者是否含0。从下标(1, 1)开始做与上个解法相同的事。最后根据两个变量来判断是否要把第一行或第一列设置全0。

A simple solution is using 2 boolean arrays whose length equal to the rows and columns of the matrix respectively to mark the presence or absence of 0 in the corresponding row or column. Then traverse all elements and check if its row/col number is marked in the corresponding position of the two arrays. Set the current element to 0 if one of both positions is marked.

An advanced solution is to use the first row and column to substitute 2 arrays to save space. However, each of them might contains 0. Therefore, 2 boolean variables are used to record if the first row and column contain 0 before we perform the same solution.

Complexity

Solution 1:

• Time complexity: $O(mn)$

• Space complexity: $O(1)$

Solution 2:

• Time complexity: $O(mn)$

• Space complexity: $O(m+n)$

Code

Solution 1: Space complexity O(1) (2 variables)

// Solution 1: Space complexity O(1) (2 variables)
func setZeroes(matrix [][]int) {
	var row0, col0 bool
	for _, l := range matrix {
		if l[0] == 0 {
			col0 = true
			break
		}
	}
	for _, v := range matrix[0] {
		if v == 0 {
			row0 = true
			break
		}
	}
	for i := 1; i < len(matrix); i++ {
		for j := 1; j < len(matrix[0]); j++ {
			if matrix[i][j] == 0 {
				matrix[i][0] = 0
				matrix[0][j] = 0
			}
		}
	}
	for i := 1; i < len(matrix); i++ {
		for j := 1; j < len(matrix[0]); j++ {
			if matrix[i][0] == 0 || matrix[0][j] == 0 {
				matrix[i][j] = 0
			}
		}
	}
	if col0 {
		for i := range matrix {
			matrix[i][0] = 0
		}
	}
	if row0 {
		for j := range matrix[0] {
			matrix[0][j] = 0
		}
	}
}

Solution 2: Space complexity O(m+n) (2 arrays)

func setZeroes(matrix [][]int) {
	rows, cols := make([]bool, len(matrix)), make([]bool, len(matrix[0]))
	for i, l := range matrix {
		for j, v := range l {
			if v == 0 {
				rows[i], cols[j] = true, true
			}
		}
	}
	for i, l := range matrix {
		for j := range l {
			if rows[i] || cols[j] {
				l[j] = 0
			}
		}
	}
}

Reference

1. 矩阵置零