Java solves the problem of special positions in binary matrices

Java solves the problem of special positions in binary matrices

01 Question

Give you a matrix of sizerows x cols, where is or , please return the number of special positions in the matrix**. matmat[i][j]01mat

Special Position Definition: Ifmat[i][j] == 1 and line i and line j All other elements in the column are 0 (row and column subscripts both start from 0 ), then Position(i, j) is called a special position.

Example 1:

输入：mat = [[1,0,0],
            [0,0,1],
            [1,0,0]]
输出：1
解释：(1,2) 是一个特殊位置，因为 mat[1][2] == 1 且所处的行和列上所有其他元素都是 0

Example 2:

输入：mat = [[1,0,0],
            [0,1,0],
            [0,0,1]]
输出：3
解释：(0,0), (1,1) 和 (2,2) 都是特殊位置

Example 3:

输入：mat = [[0,0,0,1],
            [1,0,0,0],
            [0,1,1,0],
            [0,0,0,0]]
输出：2

Example 4:

输入：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]]
输出：3

hint:

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

02 Knowledge points

• Two-dimensional array

03 My solution

public class shuzu05 {
    
    
	public static void main(String[] args) {
    
    
        //测试数据
		int[][] mat=new int[][] {
    
    
			{
    
    1,0,0},
			{
    
    0,1,0},
			{
    
    0,0,1}
		};
		System.out.println(numSpecial(mat));
	}
public static int numSpecial(int[][] mat) {
    
    
	int count=0;
	int m=mat.length;//记录排的数量
	int n=mat[0].length;//记录列的数量
	for (int i = 0; i < m; i++) {
    
    
		boolean flag=false;//用于标记是否满足第一条件
		for (int j = 0; j <n; j++) {
    
    
			if (mat[i][j]==1) {
    
    
				flag=true;
                //当满足第一条件执行，第二条件
			}
			if (flag) {
    
    
                //目的是判断同一行是否有满足第一条件的数
				for (int k1 = 0; k1 < m; k1++) {
    
    
					if (mat[k1][j]==1&&k1!=i) {
    
    
						flag=false;
						break;
					}
				}
                //目的是判断同一列是否有满足第一条件的数
				for (int k2 = 0; k2 < n; k2++) {
    
    
					if (mat[i][k2]==1&&k2!=j) {
    
    
						flag=false;
						break;//flag要修改，还要退出循环
					}

				}
   //如果前两个条件都不满足，则说明同一行和同一列都没有mat[i][j]=1,flag的值依旧是true，则count+1               
			}
			if (flag) {
    
    
				count++;
			}
		}
	}
	
	return count;
}
}