思路

方法 1：额外存储空间方法

code

public void setZeroes(int[][] matrix) {
        Set<Integer> rowSet = new HashSet<>();
        Set<Integer> columnSet = new HashSet<>();
        int rows = matrix.length;
        int columns = matrix[0].length;
        for(int i=0;i<rows;i++)
            for(int j=0;j<columns;j++){
                if(matrix[i][j]==0){
                    rowSet.add(i);      //加到集合中去
                    columnSet.add(j);
                }
            }
        for(int i=0;i<rows;i++){
            for(int j=0;j<columns;j++){
                if(rowSet.contains(i)||columnSet.contains(j)){
                    matrix[i][j]=0;   //进行更新
                }
            }
        }
    }

方法2 不使用额外空间的解法

 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;
            }
          }
        }
        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;
          }
        }
     if (isCol) {
          for (int i = 0; i < R; i++) {
            matrix[i][0] = 0;
          }
        }  
    }       
}