[leetcode] 1337. The K Weakest Rows in a Matrix

[leetcode] 1337. The K Weakest Rows in a Matrix

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 = 3Output: [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 = 2Output: [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.lengthn == mat[i].length2 <= n, m <= 1001 <= k <= mmatrix[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%​​

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