The K Weakest Rows in a Matrix
You are given an m x n binary matrix mat of 1's (representing soldiers) and 0's (representing civilians). The soldiers are positioned in front of the civilians. That is, all the 1's will appear to the left of all the 0's in each row.
A row i is weaker than a row j if one of the following is true:
The number of soldiers in row i is less than the number of soldiers in row j.
Both rows have the same number of soldiers and i < j.
Return the indices of the k weakest rows in the matrix ordered from weakest to strongest.
Example 1:
Input: mat =
[[1,1,0,0,0],
[1,1,1,1,0],
[1,0,0,0,0],
[1,1,0,0,0],
[1,1,1,1,1]],
k = 3
Output: [2,0,3]
Explanation:
The number of soldiers in each row is:
- Row 0: 2
- Row 1: 4
- Row 2: 1
- Row 3: 2
- Row 4: 5
The rows ordered from weakest to strongest are [2,0,3,1,4].
Example 2:
Input: mat =
[[1,0,0,0],
[1,1,1,1],
[1,0,0,0],
[1,0,0,0]],
k = 2
Output: [0,2]
Explanation:
The number of soldiers in each row is:
- Row 0: 1
- Row 1: 4
- Row 2: 1
- Row 3: 1
The rows ordered from weakest to strongest are [0,2,3,1].
Constraints:
m == mat.length
n == mat[i].length
2 <= n, m <= 100
1 <= k <= m
matrix[i][j] is either 0 or 1.
思路一：先计算，然后排序。用了 stream，效率比较低，可以用数字来实现，用取余来实现两个比较，但是可读性会降低
class Rows {
int numOfSoldier;
int index;
}
public int[] kWeakestRows(int[][] mat, int k) {
int[] rows = new int[mat.length];
List<Rows> list = new ArrayList<>();
for (int i = 0; i < mat.length; i++) {
int sum = 0;
for (int j = 0; j < mat[i].length; j++) {
sum += mat[i][j];
}
list.add(new Rows(sum, i));
}
return list.stream()
.sorted(Comparator.comparingInt(Rows::getNumOfSoldier).thenComparing(Rows::getIndex))
.limit(k)
.map(Rows::getIndex)
.mapToInt(Integer::intValue)
.toArray();
}