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90 lines
3.0 KiB
Markdown
90 lines
3.0 KiB
Markdown
# [1337. The K Weakest Rows in a Matrix](https://leetcode.com/problems/the-k-weakest-rows-in-a-matrix/)
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## 题目
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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.
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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*.
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**Example 1:**
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```
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Input: mat =
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[[1,1,0,0,0],
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[1,1,1,1,0],
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[1,0,0,0,0],
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[1,1,0,0,0],
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[1,1,1,1,1]],
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k = 3
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Output: [2,0,3]
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Explanation:
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The number of soldiers for each row is:
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row 0 -> 2
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row 1 -> 4
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row 2 -> 1
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row 3 -> 2
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row 4 -> 5
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Rows ordered from the weakest to the strongest are [2,0,3,1,4]
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```
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**Example 2:**
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```
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Input: mat =
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[[1,0,0,0],
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[1,1,1,1],
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[1,0,0,0],
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[1,0,0,0]],
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k = 2
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Output: [0,2]
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Explanation:
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The number of soldiers for each row is:
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row 0 -> 1
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row 1 -> 4
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row 2 -> 1
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row 3 -> 1
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Rows ordered from the weakest to the strongest are [0,2,3,1]
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```
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**Constraints:**
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- `m == mat.length`
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- `n == mat[i].length`
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- `2 <= n, m <= 100`
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- `1 <= k <= m`
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- `matrix[i][j]` is either 0 **or** 1.
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## 题目大意
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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。
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## 解题思路
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- 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。
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- 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。
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## 代码
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```go
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package leetcode
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func kWeakestRows(mat [][]int, k int) []int {
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res := []int{}
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for j := 0; j < len(mat[0]); j++ {
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for i := 0; i < len(mat); i++ {
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if mat[i][j] == 0 && ((j == 0) || (mat[i][j-1] != 0)) {
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res = append(res, i)
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}
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}
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}
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for i := 0; i < len(mat); i++ {
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if mat[i][len(mat[0])-1] == 1 {
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res = append(res, i)
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}
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}
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return res[:k]
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}
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``` |