Given a m * n matrix of distinct numbers, return all lucky numbers in the matrix in any order.

    A lucky number is an element of the matrix such that it is the minimum element in its row and maximum in its column.

    Example 1:

    1. Input: matrix = [[3,7,8],[9,11,13],[15,16,17]]
    2. Output: [15]
    3. Explanation: 15 is the only lucky number since it is the minimum in its row and the maximum in its column

    Example 2:

    1. Input: matrix = [[1,10,4,2],[9,3,8,7],[15,16,17,12]]
    2. Output: [12]
    3. Explanation: 12 is the only lucky number since it is the minimum in its row and the maximum in its column.

    Example 3:

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

    Constraints:

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

    题意

    在矩阵中,如果一个数既是它所在行的最小值,又是它所在列的最大值,则称这个数为幸运数。找到矩阵中所有的幸运数。

    思路

    每次先找到一行的最小值,再判断它是不是所在列的最大值。


    代码实现

    1. class Solution {
    2. public List<Integer> luckyNumbers (int[][] matrix) {
    3. List<Integer> ans = new ArrayList<>();
    4. if (matrix.length == 0) {
    5. return ans;
    6. }
    7. for (int i = 0; i < matrix.length; i++) {
    8. // 找到行的最小值
    9. int min = Integer.MAX_VALUE;
    10. int pos = 0;
    11. for (int j = 0; j < matrix[0].length; j++) {
    12. if (matrix[i][j] < min) {
    13. min = matrix[i][j];
    14. pos = j;
    15. }
    16. }
    17. // 判断是不是列的最大值
    18. boolean flag = true;
    19. for (int j = 0; j < matrix.length; j++) {
    20. if (matrix[j][pos] > min) {
    21. flag = false;
    22. break;
    23. }
    24. }
    25. if (flag) {
    26. ans.add(min);
    27. }
    28. }
    29. return ans;
    30. }
    31. }