# [Leetcode] 85. Maximal Rectangle Java

Source: Internet
Author: User

Topic:

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.

test instructions and analysis: is to give a matrix, find a whole is one of the largest sub-matrix. a two-dimensional matrix containing 0 and 1 is given, which requires that the maximum sub-matrix containing only 1 is obtained. Determine if it is possible to form an all-1 rectangle at the point of the matrix[i][j] and the top left corner: If the current point is 0, it is certainly not, if the current point is not 0, then how to find the top left corner can be obtained by all 1 of the largest matrix, here can be converted to the maximum area of the histogram.

Code:

`classSolution { Public intMaximalrectangle (Char[] matrix) {//a two-dimensional matrix containing 0 and 1 is given, which requires that the maximum sub-matrix containing only 1 is obtained. Determine if it is possible to form an all-1 rectangle at the point of Matrix[i][j] and the upper-left corner: If the current point is 0, it is certainly not, and if the current point is not 0,        intXlength =matrix.length; if(xlength==0)return0; intYlength = matrix[0].length; int[] Tempmatrix =New int[Xlength] [Ylength];//Tempmatrix[i][j] Represents the number of consecutive 1 in the first row of column J         for(intj=0;j<ylength;j++) {tempmatrix[0][J] = (matrix[0][j]== ' 0 '? 0:1); }         for(intj=0;j<ylength;j++) {             for(inti=1;i<xlength;i++){                if(matrix[i][j]== ' 0 ') Tempmatrix[i][j]= 0; Else{Tempmatrix[i][j]= Tempmatrix[i-1][j]+1; }            }        }        intMax = 0;  for(inti=0;i<xlength;i++){            inttemp =Largestrectanglearea (Tempmatrix[i]); if(Max <temp) Max=temp; }        returnMax; }    Private intLargestrectanglearea (int[] Heights) {//What should I do if the height array is known to be ascending? For example 1,2,5,7,8 (1*5) vs. (2*4) vs. (5*3) vs. (7*2) vs. (8*1) is Max (height[i]*), and the purpose of using stacks is to construct such ascending sequences, which are solved by the above method. Save the maximum possible value when constructing the stack        if(Heights = =NULL)            return0; intTempresult = 0; Stack<Integer> stack =NewStack<>(); Stack.push (heights[0]);  for(inti=1;i){            if(Heights[i]>=stack.peek ()) {//AscendingStack.push (Heights[i]); }Else{                if(!Stack.isempty ()) {                    intCount = 0; intMin =Stack.peek ();  while(!stack.isempty () && Stack.peek () >Heights[i]) {                        if(Stack.peek () <min) {min=Stack.peek ();                        } stack.pop (); Count++; if(tempresult<count*min) {Tempresult= count*min; }                    }                    intJ=0;  while(j<=count)                        {Stack.push (heights[i]); J++; }                }            }        }         for(inti=heights.length-1;i>=0;i--){            intx=Stack.pop (); if((heights.length-i) *x>Tempresult) {Tempresult= (heights.length-i) *x; }        }        returnTempresult; }}`

[Leetcode] 85. Maximal Rectangle Java

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