1329. Sort the Matrix Diagonally

1329. Sort the Matrix Diagonally #

题目 #

matrix diagonal is a diagonal line of cells starting from some cell in either the topmost row or leftmost column and going in the bottom-right direction until reaching the matrix’s end. For example, the matrix diagonal starting from mat[2][0], where mat is a 6 x 3 matrix, includes cells mat[2][0]mat[3][1], and mat[4][2].

Given an m x n matrix mat of integers, sort each matrix diagonal in ascending order and return the resulting matrix.

Example 1:

https://assets.leetcode.com/uploads/2020/01/21/1482_example_1_2.png

Input: mat = [[3,3,1,1],[2,2,1,2],[1,1,1,2]]
Output: [[1,1,1,1],[1,2,2,2],[1,2,3,3]]

Constraints:

  • m == mat.length
  • n == mat[i].length
  • 1 <= m, n <= 100
  • 1 <= mat[i][j] <= 100

题目大意 #

给你一个 m * n 的整数矩阵 mat ,请你将同一条对角线上的元素(从左上到右下)按升序排序后,返回排好序的矩阵。

解题思路 #

  • 这道题思路很简单。按照对角线,把每条对角线的元素读取出来放在数组中。这里可以利用 map 保存这些数组。再将这些数组排序。最后按照对角线还原矩阵即可。

代码 #

package leetcode

func diagonalSort(mat [][]int) [][]int {
	m, n, diagonalsMap := len(mat), len(mat[0]), make(map[int][]int)
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			diagonalsMap[i-j] = append(diagonalsMap[i-j], mat[i][j])
		}
	}
	for _, v := range diagonalsMap {
		sort.Ints(v)
	}
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			mat[i][j] = diagonalsMap[i-j][0]
			diagonalsMap[i-j] = diagonalsMap[i-j][1:]
		}
	}
	return mat
}

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