Maximal Rectangle leetcode java

題目：

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

 

題解：

 這道題可以應用之前解過的Largetst Rectangle in Histogram一題輔助解決。解決方法是：

按照每一行計算列中有1的個數，作為高度，當遇見0時，這一列高度就為0。然后對每一行計算 Largetst Rectangle in Histogram，最后得到的就是結果。

 

代碼如下：

 
           1  
          public  
          int maximalRectangle( 
          char[][] matrix) { 
          
 
           2          
          if(matrix== 
          null || matrix.length==0 || matrix[0].length==0) 
          
 
           3              
          return 0; 
          
 
           4          
          int m = matrix.length; 
          
 
           5          
          int n = matrix[0].length; 
          
 
           6          
          int max = 0; 
          
 
           7          
          int[] height =  
          new  
          int[n]; 
          // 
          對每一列構造數組 
          
 
           8  
                   
          for( 
          int i=0;i<m;i++){ 
          
 
           9              
          for( 
          int j=0;j<n;j++){ 
          
 
          10                  
          if(matrix[i][j] == '0') 
          // 
          如果遇見0，這一列的高度就為0了 
          
 
          11  
                              height[j] = 0; 
          
 
          12                  
          else 
          
 
          13                     height[j] += 1; 
          
 
          14             } 
          
 
          15             max = Math.max(largestRectangleArea(height),max); 
          
 
          16         } 
          
 
          17          
          return max; 
          
 
          18     } 
          
 
          19      
          
 
          20      
          public  
          int largestRectangleArea( 
          int[] height) { 
          
 
          21         Stack<Integer> stack =  
          new Stack<Integer>(); 
          
 
          22          
          int i = 0; 
          
 
          23          
          int maxArea = 0; 
          
 
          24          
          int[] h =  
          new  
          int[height.length + 1]; 
          
 
          25         h = Arrays.copyOf(height, height.length + 1); 
          
 
          26          
          while(i < h.length){ 
          
 
          27              
          if(stack.isEmpty() || h[stack.peek()] <= h[i]){ 
          
 
          28                 stack.push(i); 
          
 
          29                 i++; 
          
 
          30             } 
          else { 
          
 
          31                  
          int t = stack.pop(); 
          
 
          32                  
          int square = -1; 
          
 
          33                  
          if(stack.isEmpty()) 
          
 
          34                     square = h[t]*i; 
          
 
          35                  
          else{ 
          
 
          36                      
          int x = i-stack.peek()-1; 
          
 
          37                     square = h[t]*x; 
          
 
          38                 } 
          
 
          39                 maxArea = Math.max(maxArea, square); 
          
 
          40             } 
          
 
          41         } 
          
 
          42          
          return maxArea; 
          
 
          43     }