Description 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
Description
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 0 or 1.
分析
题目的意思是：给定一个矩阵，根据每行1的个数从小到大进行排序，我这里首先把每一行的1的个数统计出来，然后用lambda排序就能得出结果了。我的实现比较简单，就去看了一下别人的实现，发现还是先统计每一行1的个数，然后排序转换成字典，取前K个字典的key就行了。
代码
class Solution:def kWeakestRows(self, mat: List[List[int]], k: int) -> List[int]:
m=len(mat)
res=[]
for i in range(m):
num=sum(mat[i])
res.append([i,num])
return [x[0] for x in sorted(res,key=lambda x:(x[1],x[0]))][:k]
参考文献
[LeetCode] Easy Python beats 98%