[LeetCode] 1329. Sort the Matrix Diagonally 將矩陣按對(duì)角線排序

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]]

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

這道題讓給一個(gè)矩陣的對(duì)角線排序，然后返回排序后的矩陣。對(duì)角線的排序可不像矩陣的行或者列排序那么容易，想要按順序遍歷對(duì)角線絕非易事，需要很復(fù)雜的坐標(biāo)變換。這道題實(shí)際上考察了一個(gè)對(duì)角線坐標(biāo)的性質(zhì)，即處于同一條對(duì)角線上的點(diǎn)的橫縱坐標(biāo)的差均相同。這其實(shí)也不難理解，因?yàn)橥粭l對(duì)角線上的點(diǎn)可以看作是共線的，那么其斜率是相同的，則橫縱坐標(biāo)的差值一定相同。知道了這條性質(zhì)后，對(duì)于任意一個(gè)坐標(biāo)位置 (i, j)，就知道其屬于 i-j 的那條對(duì)角線，于是可以建立一個(gè)差值和其對(duì)應(yīng)的所有的點(diǎn)的映射，將同一條對(duì)角線上的所有點(diǎn)放到一個(gè)數(shù)組中，然后再對(duì)每個(gè)一個(gè)數(shù)組排序，注意這里是按從大到小排序，這樣從后往前取就是所求的順序了。之后再遍歷一次矩陣，對(duì)于每一個(gè)位置 (i, j)，到 i-j 對(duì)應(yīng)的集合中，取出當(dāng)前最后的一個(gè)數(shù)字用來(lái)更新 mat[i][j] 即可，參見(jiàn)代碼如下：

class Solution {
public:
    vector<vector<int>> diagonalSort(vector<vector<int>>& mat) {
        int m = mat.size(), n = mat[0].size();
        unordered_map<int, vector<int>> diagMap;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                diagMap[i - j].push_back(mat[i][j]);
            }
        }
        for (auto &a : diagMap) {
            sort(a.second.rbegin(), a.second.rend());
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                mat[i][j] = diagMap[i - j].back();
                diagMap[i - j].pop_back();
            }
        }
        return mat;
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1329

參考資料：

https://leetcode.com/problems/sort-the-matrix-diagonally/

https://leetcode.com/problems/sort-the-matrix-diagonally/solutions/489749/java-python-straight-forward/

https://leetcode.com/problems/sort-the-matrix-diagonally/solutions/489775/c-elegant-map-solution-easy-to-understand/

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