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 1:
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
问题简要描述:返回排好序的矩阵
细节阐述:
- 根据 j−i 的值来确定每条对角线,为了保证值为正数,加上一个偏移量 m,即 m−i+j
Java
java
class Solution {
public int[][] diagonalSort(int[][] mat) {
int m = mat.length, n = mat[0].length;
List<Integer>[] g = new List[m + n];
Arrays.setAll(g, e -> new ArrayList<>());
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
g[m - i + j].add(mat[i][j]);
}
}
for (List<Integer> e : g) {
Collections.sort(e, (a, b) -> b - a);
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
mat[i][j] = g[m - i + j].removeLast();
}
}
return mat;
}
}
Python3
python
class Solution:
def diagonalSort(self, mat: List[List[int]]) -> List[List[int]]:
m, n = len(mat), len(mat[0])
g = [[] for _ in range(m + n)]
for i, row in enumerate(mat):
for j, x in enumerate(row):
g[m - i + j].append(mat[i][j])
for e in g:
e.sort(reverse=True)
for i, row in enumerate(mat):
for j, x in enumerate(row):
mat[i][j] = g[m - i + j].pop()
return mat
TypeScript
TypeScript
function diagonalSort(mat: number[][]): number[][] {
let m = mat.length, n = mat[0].length;
let g: number[][] = Array.from({length: m + n}, () => []);
for (let i = 0; i < m; i++) {
for (let j = 0; j < n; j++) {
g[m - i + j].push(mat[i][j]);
}
}
for (const e of g) {
e.sort((a, b) => b - a);
}
for (let i = 0; i < m; i++) {
for (let j = 0; j < n; j++) {
mat[i][j] = g[m - i + j].pop();
}
}
return mat;
};