19.2.23 [LeetCode 85] Maximal Rectangle

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

题意

找图中由1组成的最大矩形

题解

class Solution {
public:
    int maximalRectangle(vector<vector<char>>& matrix) {
        if (matrix.empty())return 0;
        int m = matrix.size(), n = matrix[0].size(), ans = 0;
        vector<int>height(n, 0),left(n, 0), right(n, n);
        for (int i = 0; i < m; i++) {
            int s = 0;
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == '0') {
                    height[j] = 0;
                    left[j] = 0;
                    s = j + 1;
                }
                else {
                    left[j] = max(left[j], s);
                    height[j]++;
                }
            }
            int e = n - 1;
            for (int j = n - 1; j >= 0; j--) {
                if (matrix[i][j] == '0') {
                    right[j] = n;
                    e = j - 1;
                }
                else {
                    right[j] = min(right[j], e);
                }
            }
            for (int j = 0; j < n; j++)
                if (matrix[i][j] == '1')
                    ans = max(ans, (right[j] - left[j] + 1)*height[j]);
        }
        return ans;
    }
};

这题好难的……

思路是找到三个数组 height[x] 代表从当前行的x索引表示的元素往上数能数到多少连续的1, left[x] 表示当前行的x索引往左边数能数到多少个连续的height值>=它自己的1, right[x] 与left相仿（这些计数均包括当前索引且如果当前索引的值为0则这三个数组相应的值无意义）