[force deduction] 1337. Row K with the weakest combat effectiveness in the matrix

Give you a size of  m * n  Matrix  mat, The matrix consists of a number of soldiers and civilians , Use them separately 1 and 0 Express .

Please return to the weakest in the matrix  k  Index of rows , Sort by weakest to strongest .

If the first  i  The number of soldiers in line is less than the number of  j  That's ok , perhaps The number of soldiers in the two lines is the same, but i Less than j, So we think the i The battle effectiveness of the line is better than the first j Row weakness .

Soldiers Always The front position in a row , in other words 1 Always in 0 Before .

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]
explain ：
The number of soldiers in each line ：
That's ok 0 -> 2 
That's ok 1 -> 4 
That's ok 2 -> 1 
That's ok 3 -> 2 
That's ok 4 -> 5 
Sort these rows from the weakest to the strongest to get [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]
explain ： 
The number of soldiers in each line ：
That's ok 0 -> 1 
That's ok 1 -> 4 
That's ok 2 -> 1 
That's ok 3 -> 1 
Sort these rows from the weakest to the strongest to get [0,2,3,1]

Tips ：

m == mat.length
n == mat[i].length
2 <= n, m <= 100
1 <= k <= m
matrix[i][j] No 0 Namely 1

  Code ：

class Solution {
    public int[] kWeakestRows(int[][] mat, int k) {

        TreeMap<Integer, Integer> map = new TreeMap<>();// Ordered set 
        for (int i = 0; i < mat.length; i++) {
            int key = i;
            for (int j = 0; j < mat[i].length; j++) {
                if (mat[i][j] == 0) {
                    break;
                }
                key += mat[i][j] * 100;  //200,401,102,203,504
            }
            map.put(key, i);
        }
        int[] arr = new int[k];
        int index = 0;
        for (Integer key1 : map.keySet()) {//map.keySet() return map Medium key value 
            // It means that you will map All of the objects key Value already set The form of the collection returns , because map It's also out of order , And key Values are also non repeatable , So here we use set Collective storage key And the return also conforms to the rules .
            if (index == k) break;// Conditions at the end of the cycle 
            arr[index++] = map.get(key1);// adopt key Values obtained map Medium value, Store in arr Array 
        }
        return arr;
    }

    public static void main(String[] args) {
        Solution s = new Solution();
        int[][] arr = new int[][]{
   {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}};
        System.out.println(Arrays.toString(s.kWeakestRows(arr, 3)));
    }
}