73. Set Matrix Zeroes

Problem

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

Example 1:

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

Example 2:

Input: 
[
  [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]
]

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?

Solution

/**
 * @param {number[][]} matrix
 * @return {void} Do not return anything, modify matrix in-place instead.
 */
var setZeroes = function(matrix) {
  var m = matrix.length;
  var n = (matrix[0] || []).length;
  for (var i = 0; i < m; i++) {
    for (var j = 0; j < n; j++) {
      if (matrix[i][j] === 0) {
        left(i, j, m, n, matrix);
        right(i, j, m, n, matrix);
        up(i, j, m, n, matrix);
        down(i, j, m, n, matrix);
      } else if (matrix[i][j] === '#') {
        matrix[i][j] = 0;
      }
    }
  }
};

var left = function (i, j, m, n, matrix) {
  for (var k = j - 1; k >= 0; k--) {
    matrix[i][k] = 0;
  }
};

var right = function (i, j, m, n, matrix) {
  for (var k = j + 1; k < n; k++) {
    matrix[i][k] = matrix[i][k] === 0 ? 0 : '#';
  }
};

var up = function (i, j, m, n, matrix) {
  for (var k = i - 1; k >= 0; k--) {
    matrix[k][j] = 0;
  }
};

var down = function (i, j, m, n, matrix) {
  for (var k = i + 1; k < m; k++) {
    matrix[k][j] = matrix[k][j] === 0 ? 0 : '#';
  }
};

Explain:

把没遍历的 1 设置为 0 会影响之后的判断，先设置为 #，再改回来。

Complexity:

• Time complexity : O(m*n).
• Space complexity : O(1).