73.Set-Matrix-Zeroes

73. Set Matrix Zeroes

题目地址

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

题目描述

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?

代码

Approach 1: Additional Memory Approach

Complexity Analysis

• Time Complexity: O(M×N) where M and N are the number of rows and columns respectively.

• Space Complexity: O(M + N)

class Solution {
    public void setZeroes(int[][] matrix) {
        int R = matrix.length;
      int C = matrix[0].length;
      Set<Integer> rows = new HashSet<Integer>();
      Set<Integer> cols = new HashSet<Integer>();

      for (int i = 0; i < R; i++) {
        for (int j = 0; j < C; j++) {
          if (matrix[i][j] == 0) {
            rows.add(i);
            cols.add(j);
          }
        }
      }

      for (int i = 0; i < R; i++) {
        for (int j = 0; j < C; j++) {
          if (rows.contains(i) || cols.contains(j)) {
            matrix[i][j] = 0;
          }
        }
      }

    }
}

Approach 3: O(1) Space, Efficient Solution

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;
      }
    }

  }
}

Approach #3 Brute O(1) Space

Complexity Analysis

• Time Complexity : O((M×N)×(M+N))

• Space Complexity : O(1)

class Solution {
  public void setZeroes(int[][] matrix) {
    int MODIFIED = -1000_000;
    int R = matrix.length;
    int C = matrix[0].length;

    for (int r = 0; r < R; r++) {
      for (int c = 0; c < C; c++) {
        if (matrix[r][c] == 0) {

          for (int k = 0; k < C; k++) {
            if (matrix[r][k] != 0) {
              matrix[r][k] = MODIFIED;
            }
          }

          for (int k = 0; k < R; k++) {
            if (matrix[k][c] != 0) {
              matrix[k][c] = MODIFIED;
            }
          }

        }
      }
    }

    for (int r = 0; r < R; r++) {
      for (int c = 0; c < C; c++) {
        if (matrix[r][c] == MODIFIED) {
          matrix[r][c] = 0;
        }
      }
    }

  }
}