85. Maximal Rectangle

Problem:

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

思路：
遍历二维数组中的每一个元素，如果这个元素非0，则储存其对应的左边界和右边界，乘上对应的高度，然后求max即得到最大值。

Solution (C++):

int maximalRectangle(vector<vector<char>>& matrix) {
    if (matrix.empty() || matrix[0].empty())  return 0;
    int m = matrix.size(), n = matrix[0].size();
    vector<int> left_bound(n), right_bound(n, n-1), height(n);
    int max_area = 0;
    
    for (int i = 0; i < m; ++i) {
        int cur_left = 0, cur_right = n-1;
        for (int j = 0; j < n; ++j) {
            if (matrix[i][j] == '1')  height[j]++;
            else  height[j] = 0;
        }
        
        for (int j = 0; j < n; ++j) {
            if (matrix[i][j] == '1')  left_bound[j] = max(left_bound[j], cur_left);
            else  {left_bound[j] = 0; cur_left = j+1;}
        }
        for (int j = n-1; j >= 0; --j) {
            if (matrix[i][j] == '1')  right_bound[j] = min(right_bound[j], cur_right);
            else  {right_bound[j] = n-1; cur_right = j-1;}
        }
        for (int j = 0; j < n; ++j) {
            max_area = max(max_area, (right_bound[j]-left_bound[j]+1) * height[j]);
        }
    }
    return max_area;
}

性能：

Runtime: 16 ms  Memory Usage: 10.5 MB