1582. Special Positions in a Binary Matrix

Given a rows x cols matrix mat, where mat[i][j] is either 0 or 1, return the number of special positions in mat.

A position (i,j) is called special if mat[i][j] == 1 and all other elements in row i and column j are 0 (rows and columns are 0-indexed).

 

Example 1:

Input: mat = [[1,0,0],
              [0,0,1],
              [1,0,0]]
Output: 1
Explanation: (1,2) is a special position because mat[1][2] == 1 and all other elements in row 1 and column 2 are 0.

Example 2:

Input: mat = [[1,0,0],
              [0,1,0],
              [0,0,1]]
Output: 3
Explanation: (0,0), (1,1) and (2,2) are special positions. 

Example 3:

Input: mat = [[0,0,0,1],
              [1,0,0,0],
              [0,1,1,0],
              [0,0,0,0]]
Output: 2

Example 4:

Input: mat = [[0,0,0,0,0],
              [1,0,0,0,0],
              [0,1,0,0,0],
              [0,0,1,0,0],
              [0,0,0,1,1]]
Output: 3

 

Constraints:

• rows == mat.length
• cols == mat[i].length
• 1 <= rows, cols <= 100
• mat[i][j] is 0 or 1.

class Solution {
    public int numSpecial(int[][] mat) {
        int res = 0;
        int m = mat.length, n = mat[0].length;
        for(int i = 0; i < m; i++) {
            for(int j = 0; j < n; j++) {
                if(mat[i][j] == 1 && check(i, j, mat)) res++;
            }
        }
        return res;
    }
    public boolean check(int i, int j, int[][] mat) {
        int m = mat.length, n = mat[0].length;
        for(int k = 0; k < m; k++) {
            if(k != i && mat[k][j] == 1) return false;
        }
        for(int k = 0; k < n; k++) {
            if(k != j && mat[i][k] == 1) return false;
        }
        return true;
    }
}

或者再用一个数组记录每行列中有几个1，如果只有一个就res++

    public int numSpecial(int[][] mat) {
        int m = mat.length, n = mat[0].length, res = 0, col[] = new int[n], row[] = new int[m];
        for (int i = 0; i < m; i++) 
            for (int j = 0; j < n; j++) 
                if (mat[i][j] == 1){
                    col[j]++;
                    row[i]++;
                } 
        for (int i = 0; i < m; i++) 
            for (int j = 0; j < n; j++) 
                if (mat[i][j] == 1 && row[i] == 1 && col[j] == 1) res++;
        return res;
    }