leetcode1337. The K Weakest Rows in a Matrix

Posted 2021-03-11 seyjs

题目如下：

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.

解题思路：把每一行的1算出来比较一下就行了。

代码如下：

class Solution(object):
    def kWeakestRows(self, mat, k):
        """
        :type mat: List[List[int]]
        :type k: int
        :rtype: List[int]
        """
        val = []
        for i in range(len(mat)):
            count = 0
            for j in range(len(mat[i])):
                if mat[i][j] == 0:
                    break
                count += 1 
            val.append((i,count))
        
        def cmpf(i1,i2):
            if i1[1] != i2[1]:
                return i1[1] - i2[1]
            return i1[0] - i2[0]
        
        val.sort(cmp=cmpf)
        
        res = []
        for i in range(k):
            res.append(val[i][0])
        
        return res