1337. The K Weakest Rows in a Matrix - easy
문제
Given a m * n matrix mat of ones (representing soldiers) and zeros (representing civilians), return the indexes of the k weakest rows in the matrix ordered from the weakest to the strongest.
A row i is weaker than row j, if the number of soldiers in row i is less than the number of soldiers in row j, or they have the same number of soldiers but i is less than j. Soldiers are always stand in the frontier of a row, that is, always ones may appear first and then zeros.
제한사항
- m == mat.length
- n == mat[i].length
- 2 <= n, m <= 100
- 1 <= k <= m
- matrix[i][j] is either 0 or 1.
입출력 예
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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 for each row is:
row 0 -> 2
row 1 -> 4
row 2 -> 1
row 3 -> 2
row 4 -> 5
Rows ordered from the weakest to the strongest are [2,0,3,1,4]
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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 for each row is:
row 0 -> 1
row 1 -> 4
row 2 -> 1
row 3 -> 1
Rows ordered from the weakest to the strongest are [0,2,3,1]
풀이
- Arary
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class Solution {
public:
vector<int> kWeakestRows(vector<vector<int>>& mat, int k) {
std::vector<std::pair<int, int>> items;
std::vector<int> result;
int row_index = 0;
for (const auto& row : mat) {
int sum = std::accumulate(row.begin(), row.end(), 0);
items.push_back({sum, row_index});
++row_index;
}
std::sort(items.begin(), items.end(),
[](const std::pair<int, int>& a, const std::pair<int, int>& b){
if (a.first == b.first) {
return a.second < b.second;
}
return a.first < b.first;
}
);
for (int i = 0 ; i < k ; ++i) {
result.push_back(items[i].second);
}
return result;
}
};