"Leetcode 221" maximal Square

Given a 2D binary matrix filled with 0 's and 1 ' s, find the largest square containing all 1 's and return its area.

For example, given the following matrix:

1 0 1 0 0 1 0 1 1 1 1 1 1 1 1 1 0 0 1 0

Return 4.

Ideas:

DP, the longest side length to reach the current (not 0) node Square[i][j] = min (square[i-1][j-1], min (Square[i-1][j], square[i][j-1]) + 1, that is, the top left 3 nodes in the lowest value of one plus 1. The code looks comfortable--!

1 classSolution {2  Public:3     intMaximalsquare (vector<vector<Char>>&matrix) {4         Const introws =matrix.size ();5         if(Rows = =0)6             return 0;7         8vector<vector<Char> >::iterator it =Matrix.begin ();9         Const intColumns = (*it). Size ();Ten          One         intSquare[rows][columns], ret =0; A          -          for(inti =0; i < rows; i++) -              for(intj =0; J < columns; J + +) theSQUARE[I][J] =0; -          -          for(inti =0; i < rows; i++){ -             if(matrix[i][0] =='1') +             { -square[i][0] =1; +RET =1; A             } at         } -      -          for(inti =0; i < columns; i++){ -             if(matrix[0][i] = ='1') -             { -square[0][i] =1; inRET =1; -             } to         } +          -          for(inti =1; i < rows; i++) the         { *              for(intj =1; J < columns; J + +) $             {Panax NotoginsengSQUARE[I][J] = matrix[i][j] = ='1'? Min (square[i-1][j], min (square[i][j-1], square[i-1][j-1])) +1:0; -                  the                 if(Square[i][j] >ret) +RET =Square[i][j]; A             } the         } +         returnRET *ret; -     } $};

"Leetcode 221" maximal Square