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96c36de · Aug 26, 2020

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1380.Lucky-Numbers-in-a-Matrix

题目

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.

题目大意

给你一个 m * n 的矩阵,矩阵中的数字 各不相同 。请你按 任意 顺序返回矩阵中的所有幸运数。幸运数是指矩阵中满足同时下列两个条件的元素:

  • 在同一行的所有元素中最小
  • 在同一列的所有元素中最大

解题思路

  • 找出矩阵中的幸运数。幸运数的定义:同时满足 2 个条件,在同一行的所有元素中最小并且在同一列的所有元素中最大。
  • 简单题。按照题意遍历矩阵,找到同时满足 2 个条件的数输出即可。

代码

func luckyNumbers(matrix [][]int) []int {
	t, r, res := make([]int, len(matrix[0])), make([]int, len(matrix[0])), []int{}
	for _, val := range matrix {
		m, k := val[0], 0
		for j := 0; j < len(matrix[0]); j++ {
			if val[j] < m {
				m = val[j]
				k = j
			}
			if t[j] < val[j] {
				t[j] = val[j]
			}
		}

		if t[k] == m {
			r[k] = m
		}
	}
	for k, v := range r {
		if v > 0 && v == t[k] {
			res = append(res, v)
		}
	}
	return res
}