1329 Sort the Matrix Diagonally 将矩阵按对角线排序

Description:

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:

Example 1:

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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]]

Example 2:

Input: mat = [[11,25,66,1,69,7],[23,55,17,45,15,52],[75,31,36,44,58,8],[22,27,33,25,68,4],[84,28,14,11,5,50]]
Output: [[5,17,4,1,52,7],[11,11,25,45,8,69],[14,23,25,44,58,15],[22,27,31,36,50,66],[84,28,75,33,55,68]]

Constraints:

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

题目描述：

矩阵对角线是一条从矩阵最上面行或者最左侧列中的某个元素开始的对角线，沿右下方向一直到矩阵末尾的元素。例如，矩阵 mat 有 6 行 3 列，从 mat[2][0] 开始的矩阵对角线将会经过 mat[2][0]、mat[3][1] 和 mat[4][2] 。

给你一个 m * n 的整数矩阵 mat ，请你将同一条矩阵对角线上的元素按升序排序后，返回排好序的矩阵。

示例：

示例 1：

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输入：mat = [[3,3,1,1],[2,2,1,2],[1,1,1,2]]
输出：[[1,1,1,1],[1,2,2,2],[1,2,3,3]]

示例 2：

输入：mat = [[11,25,66,1,69,7],[23,55,17,45,15,52],[75,31,36,44,58,8],[22,27,33,25,68,4],[84,28,14,11,5,50]]
输出：[[5,17,4,1,52,7],[11,11,25,45,8,69],[14,23,25,44,58,15],[22,27,31,36,50,66],[84,28,75,33,55,68]]

提示：

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

思路：

模拟
按照对角线取出值放入优先队列中, 然后重新放回原矩阵
时间复杂度为 O(mnlgn), 空间复杂度为 O(mn)

代码:

C++:

class Solution 
{
public:
    vector<vector<int>> diagonalSort(vector<vector<int>>& mat) 
    {
        for (int k = 0, m = mat.size(), n = mat.front().size(), r = min(m, n); k < r; k++) for (int i = 1; i < m; i++) for (int j = 1; j < n; j++) if (mat\[i][j] < mat\[i - 1][j - 1]) swap(mat\[i][j], mat\[i - 1][j - 1]);
        return mat;
    }
};

Java:

class Solution {
    public int[][] diagonalSort(int[][] mat) {
        int m = mat.length, n = mat\[0].length;
        HashMap<Integer, PriorityQueue<Integer>> map = new HashMap<>();
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                map.putIfAbsent(i - j, new PriorityQueue<>());
                map.get(i - j).offer(mat\[i][j]);
            }
        }
        for (int i = 0; i < m; i++) for (int j = 0; j < n; j++) mat\[i][j] = map.get(i - j).poll();
        return mat;
    }
}

Python:

class Solution:
    def diagonalSort(self, mat: List[List[int]]) -> List[List[int]]:
        m, n, nums = len(mat), len(mat\[0]), defaultdict(list)
        for i in range(m):
            for j in range(n):
                heappush(nums[i - j], mat\[i][j])
        for i in range(m):
            for j in range(n):
                mat\[i][j] = heappop(nums[i - j])
        return mat