描述
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]
Note:
m == mat.length
n == mat[i].length
2 <= n, m <= 100
1 <= k <= m
matrix[i][j] is either 0 or 1.
解析
easy 程度的题目描述再长，都是纸脑虎，我要打十个！ 根据题意，找出兵数最少的前 k 行，那就先将 mat 中的所有行按照兵力进行从少到多的排序，然后取出前 k 行的索引即可。
解答
class Solution(object):
def kWeakestRows(self, mat, k):
"""
:type mat: List[List[int]]
:type k: int
:rtype: List[int]
"""
res = []
for i,row in enumerate(mat):
res.append((row.count(1),i))
res.sort(key=lambda x:x[0])
return[x[1] for x in res[:k]]
运行结果
Runtime: 92 ms, faster than 36.71% of Python online submissions for The K Weakest Rows in a Matrix.
Memory Usage: 13.6 MB, less than 94.30% of Python online submissions for The K Weakest Rows in a Matrix.
原题链接：leetcode.com/problems/th…
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