1380. Lucky Numbers in a Matrix

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:

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

Example 2:

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

Example 3:

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

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给定一个没有重复元素的矩阵，要求找出矩阵中的幸运数字，幸运数字的定义是它是所在行的最小值且是所在列的最大值。

有点意思的题目，如果暴力解法，遍历每个元素，同时判断它是否是所在行最小值且所在列最大值的话，时间复杂度高达$O((mn)^2)$，不可取。

更好的方法是，在遍历的过程中，记录当前行的最小值即当前列的最大值，最后判断一下最小值的数组和最大值的数组是否有相等元素，如果有的话，说明这个数是所在行的最小值且是所在列的最大值。可以这样做的原因是矩阵中没有重复元素。

完整代码如下：

class Solution {
public:
	vector<int> luckyNumbers(vector<vector<int>>& matrix) {
		vector<int> ans;
		int m = matrix.size(), n = matrix[0].size();
		vector<int> rowmin(m, INT_MAX), colmax(n, INT_MIN);
		for (int i = 0; i < m; ++i) {
			for (int j = 0; j < n; ++j) {
				rowmin[i] = min(rowmin[i], matrix[i][j]);
				colmax[j] = max(colmax[j], matrix[i][j]);
			}
		}
		for (int i = 0; i < m; ++i) {
			for (int j = 0; j < n; ++j) {
				if (rowmin[i] == colmax[j])ans.push_back(rowmin[i]);
			}
		}
		return ans;
	}
};

本代码提交AC，用时24MS。