Force deduction solution summarizes the lucky numbers in 1380 matrix

Original link ： Power button

describe ：

To give you one m * n Matrix , The number in the matrix Each are not identical . Please press arbitrarily Return all the lucky numbers in the matrix in order .

Lucky number refers to the elements in the matrix that meet the following two conditions at the same time ：

The smallest of all elements in the same row
The largest of all elements in the same column
 

Example 1：

Input ：matrix = [[3,7,8],[9,11,13],[15,16,17]]
Output ：[15]
explain ：15 Is the only lucky number , Because it is the smallest value in its row , It is also the maximum value in the column .
Example 2：

Input ：matrix = [[1,10,4,2],[9,3,8,7],[15,16,17,12]]
Output ：[12]
explain ：12 Is the only lucky number , Because it is the smallest value in its row , It is also the maximum value in the column .
Example 3：

Input ：matrix = [[7,8],[1,2]]
Output ：[7]
 

Tips ：

m == mat.length
n == mat[i].length
1 <= n, m <= 50
1 <= matrix[i][j] <= 10^5
All elements in the matrix are different

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/lucky-numbers-in-a-matrix
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

Their thinking ：

*  Their thinking ：
*  Find out the smallest one on the line , List the largest . This car will have two arrays , Repeated in these two arrays , It's a lucky number 

Code ：

public class Solution1380 {

    public List<Integer> luckyNumbers(int[][] matrix) {
        Integer[] xInts = new Integer[matrix.length];
        Integer[] yInts = new Integer[matrix[0].length];
        for (int y = 0; y < matrix.length; y++) {
            for (int x = 0; x < matrix[0].length; x++) {
                int value = matrix[y][x];
                if (xInts[y] == null) {
                    xInts[y] = value;
                } else {
                    //x Axis 
                    xInts[y] = Math.min(xInts[y], value);
                }
                if (yInts[x] == null) {
                    yInts[x] = value;
                } else {
                    //y Axis 
                    yInts[x] = Math.max(yInts[x], value);
                }
            }
        }
        List<Integer> result = new ArrayList<>();
        Set<Integer> cache = new HashSet<>(Arrays.asList(xInts));
        for (Integer i : yInts) {
            if (cache.contains(i)) {
                result.add(i);
            }
        }
        return result;
    }
}