1341. The K Weakest Rows in a Matrix

Posted 2021-03-12 wentiliangkaihua

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.

class Solution {
    public int[] kWeakestRows(int[][] mat, int k) {
        int le = mat.length;
        int[][] help = new int[le][2];
        
        for(int i = 0; i < le; i++){
            help[i][1] = helper(mat[i]);
            help[i][0] = i;
        }
        Arrays.sort(help, (a, b) -> a[1] - b[1]);
        int t = k;
        int[] res = new int[k];
        for(int i = 0; i < k; i++){
            res[i] = help[i][0];
        }
        return res;
    }
    public int helper(int[] arr){
        int res = 0;
        for(int i: arr) res = i == 1 ? res + 1 : res;
        return res;
    }
}

lambda表达式??‍