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

題意

找圖中由1組成的最大矩形

題解

 1 class Solution {
 2 public
 
:
 3     int maximalRectangle(vector<vector<char>>& matrix) {
 4         if (matrix.empty())return 0;
 5         int m = matrix.size(), n = matrix[0].size(), ans = 0;
 6         vector<int>height(n, 0),left(n, 0), right(n, n);
 7         for (int i = 0; i < m; i++) {
 8             int
 
 s = 0;
 9             for (int j = 0; j < n; j++) {
10                 if (matrix[i][j] == ‘0‘) {
11                     height[j] = 0;
12                     left[j] = 0;
13                     s = j + 1;
14                 }
15                 else {
16                     left[j] = max(left[j], s);
 
17                     height[j]++;
18                 }
19             }
20             int e = n - 1;
21             for (int j = n - 1; j >= 0; j--) {
22                 if (matrix[i][j] == ‘0‘) {
23                     right[j] = n;
24                     e = j - 1;
25                 }
26                 else {
27                     right[j] = min(right[j], e);
28                 }
29             }
30             for (int j = 0; j < n; j++)
31                 if (matrix[i][j] == ‘1‘)
32                     ans = max(ans, (right[j] - left[j] + 1)*height[j]);
33         }
34         return ans;
35     }
36 };

這題好難的……

思路是找到三個數組 height[x] 代表從當前行的x索引表示的元素往上數能數到多少連續的1, left[x] 表示當前行的x索引往左邊數能數到多少個連續的height值>=它自己的1, right[x] 與left相仿（這些計數均包括當前索引且如果當前索引的值為0則這三個數組相應的值無意義）

19.2.23 [LeetCode 85] Maximal Rectangle