Leetcode Maximal Rectangle-----java

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

For example, given the following matrix:

1 1 1 1 1 11 0 0 1 0

Return 6.

The largest sub-matrix area consisting of 1 in the matrix is evaluated.

1, first to observe, found that if the direct traversal and then find the maximum matrix of each point is obviously wrong, time complexity is too high.

2, combined with 84 questions, found that the two problems are very similar to the problem given in the example can be carefully observed can be found that the maximum sub-matrix is a line as the bottom, and then with the method of 84 questions to solve, so first to the matrix conversion, the matrix can be converted to the following:

1 0 1 0 0 1 0 1 1 11 1 1 1 1 1 0 0 1 0

After conversion:

1 0 1 0 02 0 2 1 13 1 3 2 24 0 0 3 0

The maximum matrix area is then solved for each row.

 Public classSolution { Public intMaximalrectangle (Char[] matrix) {        intLen1 =matrix.length; if(Len1 = = 0)            return0; intLen2 = matrix[0].length; if(Len2 = = 0)            return0; intresult = 0; int[] Heights =New int[Len1][len2];  for(inti = 0;i<len2;i++){            if(Matrix[0][i] = = ' 1 ') heights[0][i] = 1; }         for(inti = 1;i < len1;i++){             for(intj = 0;j<len2;j++){                if(Matrix[i][j] = = ' 1 ') Heights[i][j]= Heights[i-1][j]+1; }        }         for(inti = 0;i<len1;i++) {result=Math.max (Result,largestrectanglearea (heights[i)); }        returnresult; }     Public intLargestrectanglearea (int[] Heights) {        intLen =heights.length; intresult = 0; if(len = = 0)            return0; int[] left =New int[Len]; int[] right =New int[Len]; left[0] = 0;  for(inti = 1;i<len;i++){            intCurleft = i-1;  while(curleft >= 0 && heights[curleft]>=Heights[i]) {Curleft= Left[curleft]-1; } Left[i]= Curleft+1; } Right[len-1] = len-1;  for(inti = len-2;i>=0;i--){            intCurright = i+1;  while(Curright<len && heights[curright]>=Heights[i]) Curright= Right[curright]+1; Right[i]= CurRight-1; }         for(inti = 0;i<len;i++) {result= Math.max (result, (right[i]-left[i]+1) *(Heights[i])); }        returnresult;}}

Leetcode maximal Rectangle-----java