Leetcode(85)maximal-rectangle（最大矩形面积）

题目描述：

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

解题思路：

1、动态规划：

 

class Solution {
    public int maximalRectangle(char[][] matrix) {
        if(matrix == null || matrix.length == 0)
            return 0;
        int rows = matrix.length;
        int cols = matrix[0].length;
        int[][] dp = new int[rows][cols];
        int area = 0;
        for(int i = 0; i < rows; i++) {
            for(int j = 0; j < cols; j++) {   
                if(i == 0) {
                    dp[i][j] = matrix[i][j] == '1' ? 1 : 0;
                    
                }else {
                    dp[i][j] = matrix[i][j] == '1' ? (dp[i - 1][j] + 1) : 0;
                }
                int min = dp[i][j];
                for(int k = j; k >= 0; k--) {
                    if(min == 0) break;
                    if(dp[i][k] < min) min = dp[i][k];  
                    area = Math.max(area, min * (j - k + 1));
                }
                
            }
        }
        return area;
    }
   
}

2、遍历标记

class Solution {
    public int maximalRectangle(char[][] matrix) {
        if(matrix == null || matrix.length == 0)
            return 0;
        int rows = matrix.length;
        int cols = matrix[0].length;
        int area = 0;
        for(int i = 0; i < rows; i++) {
            for(int j = 0; j < cols; j++) {   
                int size = maxArea(matrix, i, j);
                area = Math.max(size, area);          
            }
        }
        return area;
    }
    public int maxArea(char[][] matrix, int m, int n) {
        int area = 0;
        int minSize = Integer.MAX_VALUE;
        for(int i = m; i < matrix.length && matrix[i][n] == '1'; i++) {
            int size = 0;
            for(int j = n; j < matrix[0].length && matrix[i][j] == '1'; j++) {
                size++;
                
            }
            minSize = Math.min(minSize, size);
            area = Math.max(area, minSize * (i - m + 1));
            
        }
        return area;   
    }
}