You are given a binary matrix matrix of size m x n, and you are allowed to rearrange the columns of the matrix in any order.
Return the area of the largest submatrix within matrix where every element of the submatrix is 1 after reordering the columns optimally.
Example 1:
Input: matrix = [[0,0,1],[1,1,1],[1,0,1]]
Output: 4
Explanation: You can rearrange the columns as shown above.
The largest submatrix of 1s, in bold, has an area of 4.
Example 2:
Input: matrix = [[1,0,1,0,1]]
Output: 3
Explanation: You can rearrange the columns as shown above.
The largest submatrix of 1s, in bold, has an area of 3.
Example 3:
Input: matrix = [[1,1,0],[1,0,1]]
Output: 2
Explanation: Notice that you must rearrange entire columns, and there is no way to make a submatrix of 1s larger than an area of 2.
Constraints:
m == matrix.length
n == matrix[i].length
1 <= m * n <= 105
matrix[i][j] is either 0 or 1.
重新排列后的最大子矩阵。
给你一个二进制矩阵 matrix ，它的大小为 m x n ，你可以将 matrix 中的 列 按任意顺序重新排列。
请你返回最优方案下将 matrix 重新排列后，全是 1 的子矩阵面积。
思路
类似85题，我们先从上到下记录矩阵里每一列出现的连续的 1 的个数，比如第一个例子，matrix = [[0,0,1],[1,1,1],[1,0,1]]，从上到下记录之后，矩阵变成
matrix = [
[0,0,1],
[1,1,2],
[2,0,3]
]
然后我们对这个矩阵逐行扫描，看一下如果以当前行为结尾，能组成的最大子矩阵的面积是多少。这样一来这道题就跟85题很像了。
复杂度
时间O(mn)
空间O(1)
代码
Java实现
class Solution {
public int largestSubmatrix(int[][] matrix) {
// corner case
if (matrix == null || matrix.length == 0) {
return 0;
}
// normal case
int m = matrix.length;
int n = matrix[0].length;
for (int i = 1; i < m; i++) {
for (int j = 0; j < n; j++) {
if (matrix[i][j] == 1) {
matrix[i][j] += matrix[i - 1][j];
}
}
}
int res = 0;
for (int i = 0; i < m; i++) {
Arrays.sort(matrix[i]);
for (int j = n - 1; j >= 0; j--) {
int area = matrix[i][j] * (n - j);
res = Math.max(res, area);
}
}
return res;
}
}
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