如何找到由四个相同角落组成的二维阵列中最大的矩形？

Question

考虑这个数组:

[
  ["B", "C", "C", "C", "C", "B", "B", "C", "A", "A"],
  ["B", "A", "C", "B", "B", "A", "B", "B", "A", "A"],
  ["B", "C", "B", "C", "A", "A", "A", "B", "C", "B"],
  ["B", "B", "B", "A", "C", "B", "A", "C", "B", "A"],
  ["A", "A", "A", "C", "A", "C", "C", "B", "A", "C"],
  ["A", "B", "B", "A", "A", "C", "B", "C", "C", "C"],
  ["C", "B", "A", "A", "C", "B", "B", "C", "A", "A"]
]

我试图获得这个2D数组中面积最大的矩形的宽度和高度.答案应该是8*4 = 32(坐标(1,1),(1,8),(4,1)和(4,8)),因为它具有相同角的最大区域"A".

Answer 1

只是因为很有趣：

var l = [ 
["B", "C", "C", "C", "C", "B", "B", "C", "A", "A"],
["B", "A", "C", "B", "B", "A", "B", "B", "A", "A"],
["B", "C", "B", "C", "A", "A", "A", "B", "C", "B"],
["B", "B", "B", "A", "C", "B", "A", "C", "B", "A"],
["A", "A", "A", "C", "A", "C", "C", "B", "A", "C"],
["A", "B", "B", "A", "A", "C", "B", "C", "C", "C"],
["C", "B", "A", "A", "C", "B", "B", "C", "A", "A"]];

var squares = [];
//for each position
for (var column=0; column< l[0].length; column++) 
   for (var row=0; row< l.length ; row++ ) 
    //look for all scuares from this position:
    for (var next_column = column+1; next_column < l[0].length; next_column++ )
        if ( l[row][next_column] == l[row][column] )
            for (var next_row = row+1; next_row < l.length; next_row++)
               //if it is a square
               if (l[next_row][column] == l[row][column] && 
                   l[next_row][next_column] == l[row][column]) 
                    //annotate square
                    squares.push( [ column, row, next_column, next_row  ] );

//get areas from squares
var area_l = squares.map(function(an_square) { 
    return (an_square[2]-an_square[0]+1)*(an_square[3]-an_square[1]+1);
  });

//search for big area
var max_area_index = area_l.indexOf(Math.max(  ...area_l  ));

//here it is
console.log( squares[max_area_index] );

结果：数组(4) [1, 1, 8, 4]