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Formatted question description: https://leetcode.ca/all/1329.html
1329. Sort the Matrix Diagonally (Medium)
A 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:
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
Solution 1.
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class Solution { public int[][] diagonalSort(int[][] mat) { int rows = mat.length, columns = mat[0].length; int startRow = rows - 2, startColumn = 0; while (startRow > 0) { List<Integer> diagonalList = new ArrayList<Integer>(); for (int i = startRow, j = startColumn; i < rows && j < columns; i++, j++) diagonalList.add(mat[i][j]); Collections.sort(diagonalList); for (int i = startRow, j = startColumn, index = 0; i < rows && j < columns; i++, j++, index++) mat[i][j] = diagonalList.get(index); startRow--; } if (startRow < 0) return mat; while (startColumn < columns) { List<Integer> diagonalList = new ArrayList<Integer>(); for (int i = startRow, j = startColumn; i < rows && j < columns; i++, j++) diagonalList.add(mat[i][j]); Collections.sort(diagonalList); for (int i = startRow, j = startColumn, index = 0; i < rows && j < columns; i++, j++, index++) mat[i][j] = diagonalList.get(index); startColumn++; } return mat; } } ############ class Solution { public int[][] diagonalSort(int[][] mat) { int m = mat.length, n = mat[0].length; for (int k = 0; k < Math.min(m, n) - 1; ++k) { for (int i = 0; i < m - 1; ++i) { for (int j = 0; j < n - 1; ++j) { if (mat[i][j] > mat[i + 1][j + 1]) { int t = mat[i][j]; mat[i][j] = mat[i + 1][j + 1]; mat[i + 1][j + 1] = t; } } } } return mat; } }
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// OJ: https://leetcode.com/problems/sort-the-matrix-diagonally/ // Time: O((M + N) * XlogX) where X is min(M, N) // Space: O(X) class Solution { public: vector<vector<int>> diagonalSort(vector<vector<int>>& A) { int M = A.size(), N = A[0].size(); for (int i = 0; i < M; ++i) { vector<int> v; for (int x = i, y = 0; x < M && y < N; ++x, ++y) v.push_back(A[x][y]); sort(begin(v), end(v)); for (int x = i, y = 0; x < M && y < N; ++x, ++y) A[x][y] = v[y]; } for (int j = 1; j < N; ++j) { vector<int> v; for (int x = 0, y = j; x < M && y < N; ++x, ++y) v.push_back(A[x][y]); sort(begin(v), end(v)); for (int x = 0, y = j; x < M && y < N; ++x, ++y) A[x][y] = v[x]; } return A; } };
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class Solution: def diagonalSort(self, mat: List[List[int]]) -> List[List[int]]: m, n = len(mat), len(mat[0]) for k in range(min(m, n) - 1): for i in range(m - 1): for j in range(n - 1): if mat[i][j] > mat[i + 1][j + 1]: mat[i][j], mat[i + 1][j + 1] = mat[i + 1][j + 1], mat[i][j] return mat
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func diagonalSort(mat [][]int) [][]int { m, n := len(mat), len(mat[0]) for k := 0; k < m-1 && k < n-1; k++ { for i := 0; i < m-1; i++ { for j := 0; j < n-1; j++ { if mat[i][j] > mat[i+1][j+1] { mat[i][j], mat[i+1][j+1] = mat[i+1][j+1], mat[i][j] } } } } return mat }