題目：

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

Example:

Input:
[
  ["1","0","1","0","0"],
  ["1","0","1","1","1"],
  ["1","1","1","1","1"],
  ["1","0","0","1","0"]
]
Output: 6

分析：

class Solution {
    public int maximalRectangle(char[][] matrix) {
        //給定矩陣座標，其中求出最大面積的表示
        //採用分別記錄左右高可達邊界座標，從而面積值為（右-左）*gao
        if(matrix==null||matrix.length==0) return 0;
        int m=matrix.length,n=matrix[0].length;
        int [] right=new int[n];
        int [] left=new int[n];
        int [] height=new int[n];
        int maxArea=0;
        
        //對於右邊界預設值為n
        Arrays.fill(right,n);
        for(int i=0;i<m;i++){
            int curLeft=0;
            int curRight=n;
            //求高
            for(int j=0;j<n;j++){
                if(matrix[i][j]=='1'){
                    height[j]++;
                }else{
                    height[j]=0;
                }
            }
            
            //求左邊界
            for(int j=0;j<n;j++){
                if(matrix[i][j]=='1'){
                    left[j]=Math.max(left[j],curLeft);
                }else{
                    left[j]=0;
                    curLeft=j+1;
                }
            }
            
            //求右邊界(從右邊開始)
            for(int j=n-1;j>=0;j--){
                if(matrix[i][j]=='1'){
                    right[j]=Math.min(right[j],curRight);
                }else{
                    right[j]=n;
                    curRight=j;
                }
            }
            
            
            //求面積
            for(int j=0;j<n;j++){
                maxArea=Math.max(maxArea,(right[j]-left[j])*height[j]);
            }
        }
        return maxArea;
       
    }
}