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。然後對每一行計算 Largetst Rectangle in Histogram，最後得到的就是結果。

 

程式碼如下：

 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];//對每一列構造陣列
 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     }