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.
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]
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]
Constraints:
-
m == mat.length -
n == mat[i].length -
2 <= n, m <= 100 -
1 <= k <= m -
matrix[i][j] is either 0or1.
题解：
class Solution {
public:
static bool cmp(pair<int, int> a, pair<int, int> b) {
if (a.second != b.second) {
return a.second < b.second;
}
return a.first < b.first;
}
vector<int> kWeakestRows(vector<vector<int>>& mat, int k) {
int m = mat.size(), n = mat[0].size();
vector<pair<int, int>> num;
for (int i = 0; i < m; i++) {
int cnt = 0;
for (int j = 0; j < n; j++) {
if (mat[i][j] == 1) {
cnt++;
}
}
num.push_back({i, cnt});
}
sort(num.begin(), num.end(), cmp);
vector<int> res;
for (int i = 0; i < k; i++) {
res.push_back(num[i].first);
}
return res;
}
};
