[LeetCode] 73. Set Matrix Zeroes

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?

Example 1:

Input: matrix = [[1,1,1],[1,0,1],[1,1,1]]
Output: [[1,0,1],[0,0,0],[1,0,1]]

Example 2:

Input: matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]
Output: [[0,0,0,0],[0,4,5,0],[0,3,1,0]]

Constraints:

• m == matrix.length
• n == matrix[0].length
• 1 <= m, n <= 200
• -231 <= matrix[i][j] <= 231 - 1

矩阵置零。

给定一个 m x n 的矩阵，如果一个元素为 0 ，则将其所在行和列的所有元素都设为 0 。请使用 原地 算法。

数组类型的题有好几道都明确要求不能使用额外空间，这题也不例外。因为不能用到额外空间，所以只能将是否需要赋零的flag记录在input的数组里面，这里我选择记录在第一行和第一列上。但是我此时仍然需要两个flag来分别记录第一行和第一列上有没有0。因为第一行和第一列上的0有可能是因着记录里面的行和列的结果而被赋值成0的。我参考了grandyang大神的思路，非常简洁，列在这里。

• 先扫描第一行第一列，如果有0，则将各自的flag设置为true
• 然后扫描除去第一行第一列的整个数组，如果有0，则将对应的第一行和第一列的数字赋0
• 再次遍历除去第一行第一列的整个数组，如果对应的第一行和第一列的数字有一个为0，则将当前值赋0
• 最后根据第一行第一列的flag来更新第一行第一列

时间O(mn)

空间O(1)

JavaScript实现

/**
 * @param {number[][]} matrix
 * @return {void} Do not return anything, modify matrix in-place instead.
 */
var setZeroes = function (matrix) {
    // corner case
    if (matrix.length === 0 || matrix === null) return;

    // normal case
    let m = matrix.length
    let n = matrix[0].length;
    let rowZero = false;
    let colZero = false;
    // check if first row has zero
    for (let i = 0; i < m; i++) {
        if (matrix[i][0] === 0) colZero = true;
    }

    // check if first column has zero
    for (let i = 0; i < n; i++) {
        if (matrix[0][i] === 0) rowZero = true;
    }

    // scan all the other rows and columns
    for (let i = 1; i < m; i++) {
        for (let j = 1; j < n; j++) {
            if (matrix[i][j] === 0) {
                matrix[0][j] = 0;
                matrix[i][0] = 0;
            }
        }
    }

    // scan again to see if we should mark the whole row or column to be zero
    for (let i = 1; i < m; i++) {
        for (let j = 1; j < n; j++) {
            if (matrix[i][0] === 0 || matrix[0][j] === 0) {
                matrix[i][j] = 0;
            }
        }
    }

    // check if we should mark the first row and first column to be zero
    if (rowZero) {
        for (let i = 0; i < n; i++) matrix[0][i] = 0;
    }
    if (colZero) {
        for (let i = 0; i < m; i++) matrix[i][0] = 0;
    }
};

 

Java实现

class Solution {
    public void setZeroes(int[][] matrix) {
        // corner case
        if (matrix == null || matrix.length == 0) {
            return;
        }
        
        // normal case
        int m = matrix.length;
        int n = matrix[0].length;
        boolean row = false;
        boolean col = false;
        // mark if first row and first col has zero
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == 0) {
                    matrix[i][0] = matrix[0][j] = 0;
                    if (i == 0) row = true;
                    if (j == 0) col = true;
                }
            }
        }
        
        // change the inner part to zeroes based on the first row and first col
        for (int i = 1; i < m; i++) {
            if (matrix[i][0] == 0) {
                for (int j = 1; j < n; j++) {
                    matrix[i][j] = 0;
                }
            }
        }
        for (int j = 1; j < n; j++) {
            if (matrix[0][j] == 0) {
                for (int i = 1; i < m; i++) {
                    matrix[i][j] = 0;
                }
            }
        }
        
        // change the first row and first col if needed
        if (row) {
            for (int j = 0; j < n; j++) {
                matrix[0][j] = 0;
            }
        }
        if (col) {
            for (int i = 0; i < m; i++) {
                matrix[i][0] = 0;
            }
        }
    }
}

 

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