1337. The K Weakest Rows in a Matrix #
题目 #
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.
题目大意 #
给你一个大小为 m * n 的矩阵 mat,矩阵由若干军人和平民组成,分别用 1 和 0 表示。请你返回矩阵中战斗力最弱的 k 行的索引,按从最弱到最强排序。如果第 i 行的军人数量少于第 j 行,或者两行军人数量相同但 i 小于 j,那么我们认为第 i 行的战斗力比第 j 行弱。军人 总是 排在一行中的靠前位置,也就是说 1 总是出现在 0 之前。
解题思路 #
- 简单题。第一个能想到的解题思路是,先统计每一行 1 的个数,然后将结果进行排序,按照 1 的个数从小到大排序,如果 1 的个数相同,再按照行号从小到大排序。排好序的数组取出前 K 位即为答案。
- 此题还有第二种解法。在第一种解法中,并没有用到题目中“军人 总是 排在一行中的靠前位置,也就是说 1 总是出现在 0 之前。”这一条件。由于有了这个条件,使得如果按照列去遍历,最先出现 0 的行,则是最弱的行。行号小的先被遍历到,所以相同数量 1 的行,行号小的会排在前面。最后记得再添加上全 1 的行。同样,最终输出取出前 K 位即为答案。此题解法二才是最优雅最高效的解法。
代码 #
package leetcode
func kWeakestRows(mat [][]int, k int) []int {
res := []int{}
for j := 0; j < len(mat[0]); j++ {
for i := 0; i < len(mat); i++ {
if mat[i][j] == 0 && ((j == 0) || (mat[i][j-1] != 0)) {
res = append(res, i)
}
}
}
for i := 0; i < len(mat); i++ {
if mat[i][len(mat[0])-1] == 1 {
res = append(res, i)
}
}
return res[:k]
}