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【LAMMPS学习】八、基础知识(6.3)使用 LAMMPS GUI

年夜雪 2024-05-13 阅读 36

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:

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

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

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;    
};
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