1337. The weakest K row in the matrix

Give you a m * n matrix of  size  mat, the matrix is ​​composed of a number of soldiers and civilians, represented by 1 and 0 respectively.

Please return k the index of the row with the weakest combat effectiveness in the matrix  , sorted from weakest to strongest.

If the first  i  the number of soldiers line is less than the first  j  line, two lines or the same number of soldiers but  i  is less than  j , then we think the first  i  fighting than the first row  j  weak line.

Soldiers  are always  at the top of the line, which means that 1 always appears before 0.

 

Example 1:

输入：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
输出：[2,0,3]
解释：
每行中的军人数目：
行 0 -> 2 
行 1 -> 4 
行 2 -> 1 
行 3 -> 2 
行 4 -> 5 
从最弱到最强对这些行排序后得到 [2,0,3,1,4]

Example 2:

输入：mat = 
[[1,0,0,0],
 [1,1,1,1],
 [1,0,0,0],
 [1,0,0,0]], 
k = 2
输出：[0,2]
解释： 
每行中的军人数目：
行 0 -> 1 
行 1 -> 4 
行 2 -> 1 
行 3 -> 1 
从最弱到最强对这些行排序后得到 [0,2,3,1]

 

prompt:

• m == mat.length
• n == mat[i].length
• 2 <= n, m <= 100
• 1 <= k <= m
• matrix[i][j] Either 0 or 1

import java.util.Arrays;

public class Solution1337 {
	public int[] kWeakestRows(int[][] mat, int k) {
		int[] out = new int[k];
		int[] sum = new int[mat.length];
		int count = 0;
		for (int i = 0; i < mat.length; i++) {
			count = 0;
			for (int j = 0; j < mat[0].length; j++) {
				if (mat[i][j] == 1) {
					count++;
				}
			}
			sum[i] = count;
		}
		// System.out.println(Arrays.toString(sum));
		int[] sort = sum.clone();
		Arrays.sort(sort);
		// System.out.println(Arrays.toString(sort));
		for (int i = 0; i < k; i++) {
			for (int j = 0; j < sum.length; j++) {
				if (sort[i] == sum[j]) {
					sum[j] = -1;
					out[i] = j;
					break;
				}
			}
		}
		return out;
	}

	public static void main(String[] args) {

		Solution1337 sol = new Solution1337();

		int[][] 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 } };
		int k = 3;
		System.out.println(Arrays.toString(sol.kWeakestRows(mat, k)));
	}
}