leecode_73. Set Matrix Zeroes

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?

将二维数组中0所在行、所在列都置为0。

我的不优秀的想法(￣ ‘i ￣;)：用两个Set分别存放要被置为0的行和列，最后遍历Set.

class Solution {
    public void setZeroes(int[][] matrix) {
        HashSet<Integer> rowSet = new HashSet<>(), colSet = new HashSet<>();
        for (int i = 0; i < matrix.length; i++){
            for (int j = 0; j < matrix[i].length; j++){
                if (matrix[i][j] == 0){
                    rowSet.add(i);
                    colSet.add(j);
                }
            }
        }
        for (int i = 0; i < Math.max(matrix.length, matrix[0].length); i++){
            if (rowSet.contains(i)) {
                for (int j = 0; j < matrix[i].length; j++) {
                    matrix[i][j] = 0;
                }
                rowSet.remove(i);
            }
            if (colSet.contains(i)){
                for (int j = 0; j < matrix.length; j++)
                    matrix[j][i] = 0;
                colSet.remove(i);
            }
        }
    }
}

解法（leetcode官方）：遍历到0时，把行首和列首置为0，继续遍历。全部遍历后，再次从头遍历，如果行首是0，则该行都置0，如果列首是0，则该列全置0.

class Solution {
  public void setZeroes(int[][] matrix) {
    Boolean isCol = false;
    int R = matrix.length;
    int C = matrix[0].length;

    for (int i = 0; i < R; i++) {

      // Since first cell for both first row and first column is the same i.e. matrix[0][0]
      // We can use an additional variable for either the first row/column.
      // For this solution we are using an additional variable for the first column
      // and using matrix[0][0] for the first row.
      if (matrix[i][0] == 0) {
        isCol = true;
      }

      for (int j = 1; j < C; j++) {
        // If an element is zero, we set the first element of the corresponding row and column to 0
        if (matrix[i][j] == 0) {
          matrix[0][j] = 0;
          matrix[i][0] = 0;
        }
      }
    }

    // Iterate over the array once again and using the first row and first column, update the elements.
    for (int i = 1; i < R; i++) {
      for (int j = 1; j < C; j++) {
        if (matrix[i][0] == 0 || matrix[0][j] == 0) {
          matrix[i][j] = 0;
        }
      }
    }

    // See if the first row needs to be set to zero as well
    if (matrix[0][0] == 0) {
      for (int j = 0; j < C; j++) {
        matrix[0][j] = 0;
      }
    }

    // See if the first column needs to be set to zero as well
    if (isCol) {
      for (int i = 0; i < R; i++) {
        matrix[i][0] = 0;
      }
    }
  }
}