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1253. Reconstruct a 2-Row Binary Matrix
Given the following details of a matrix with n columns and 2 rows :
-
The matrix is a binary matrix, which means each element in the matrix can be
0or1. -
The sum of elements of the 0-th(upper) row is given as
upper. -
The sum of elements of the 1-st(lower) row is given as
lower. -
The sum of elements in the i-th column(0-indexed) is
colsum[i], wherecolsumis given as an integer array with lengthn.
Your task is to reconstruct the matrix with upper, lower and colsum.
Return it as a 2-D integer array.
If there are more than one valid solution, any of them will be accepted.
If no valid solution exists, return an empty 2-D array.
Example 1:
Input: upper = 2, lower = 1, colsum = [1,1,1]
Output: [[1,1,0],[0,0,1]]
Explanation: [[1,0,1],[0,1,0]], and [[0,1,1],[1,0,0]] are also correct answers.Example 2:
Input: upper = 2, lower = 3, colsum = [2,2,1,1]
Output: []Example 3:
Input: upper = 5, lower = 5, colsum = [2,1,2,0,1,0,1,2,0,1]
Output: [[1,1,1,0,1,0,0,1,0,0],[1,0,1,0,0,0,1,1,0,1]]Constraints:
-
1 <= colsum.length <= 10^5 -
0 <= upper, lower <= colsum.length -
0 <= colsum[i] <= 2
思路分析
为什么需要现在 upper 和 lower 中先减去 2 的数量呢？没想明白
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-09-25 21:12:25
*/
public List<List<Integer>> reconstructMatrix(int upper, int lower, int[] colsum) {
int sum = 0;
int twos = 0;
for (int num : colsum) {
sum += num;
if (num == 2) {
twos++;
}
}
if (sum != upper + lower || upper < twos || lower < twos) {
return Collections.emptyList();
}
List<List<Integer>> result = new ArrayList<>(2);
result.add(new ArrayList<>(colsum.length));
result.add(new ArrayList<>(colsum.length));
upper -= twos;
lower -= twos;
for (int i = 0; i < colsum.length; i++) {
int up = 0;
int low = 0;
if (colsum[i] == 2) {
up = 1;
low = 1;
} else if (colsum[i] == 1) {
if (upper > 0) {
up = 1;
upper--;
} else {
low = 1;
lower--;
}
}
result.get(0).add(up);
result.get(1).add(low);
}
return result;
}