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The K Weakest Rows in a Matrix (E)

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.

题意

给定一个二维数组，将其按照每个一维数组中1的个数从小到大排序，取前K个。

思路

直接建堆排序处理。

代码实现

Java

class Solution {
    public int[] kWeakestRows(int[][] mat, int k) {
        Queue<int[]> heap = new PriorityQueue<>(k, (a, b) -> a[1] - b[1] != 0 ? a[1] - b[1] : a[0] - b[0]);
        for (int i = 0; i < mat.length; i++) {
            int cnt = 0;
            for (int j = 0; j < mat[i].length; j++) {
                if (mat[i][j] == 1) cnt++;
            }
            heap.offer(new int[]{i, cnt});
        }
        int[] ans = new int[k];
        for (int i = 0; i < k; i++) {
            ans[i] = heap.poll()[0];
        }
        return ans;
    }
}

1337. The K Weakest Rows in a Matrix (E)

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原文地址：https://www.cnblogs.com/mapoos/p/14404290.html