Leetcode 1030 Matrix Cells in Distance Order (排序)

Posted 2020-11-25 Will

Leetcode 1030

题目描述

We are given a matrix with R rows and C columns has cells with integer coordinates (r, c), where 0 <= r < R and 0 <= c < C.

Additionally, we are given a cell in that matrix with coordinates (r0, c0).

Return the coordinates of all cells in the matrix, sorted by their distance from (r0, c0) from smallest distance to largest distance.  Here, the distance between two cells (r1, c1) and (r2, c2) is the Manhattan distance, |r1 - r2| + |c1 - c2|.  (You may return the answer in any order that satisfies this condition.)

例子

Example 1:
Input: R = 1, C = 2, r0 = 0, c0 = 0
Output: [[0,0],[0,1]]
Explanation: The distances from (r0, c0) to other cells are: [0,1]

Example 2:
Input: R = 2, C = 2, r0 = 0, c0 = 1
Output: [[0,1],[0,0],[1,1],[1,0]]
Explanation: The distances from (r0, c0) to other cells are: [0,1,1,2]
The answer [[0,1],[1,1],[0,0],[1,0]] would also be accepted as correct.

Example 3:
Input: R = 2, C = 3, r0 = 1, c0 = 2
Output: [[1,2],[0,2],[1,1],[0,1],[1,0],[0,0]]
Explanation: The distances from (r0, c0) to other cells are: [0,1,1,2,2,3]
There are other answers that would also be accepted as correct, such as [[1,2],[1,1],[0,2],[1,0],[0,1],[0,0]].

解题思路

新建一个Array，对各个元素按abs(r-r0)+abs(c-c0)排序即可。

代码

Python3
class Solution:
    def allCellsDistOrder(self, R: int, C: int, r0: int, c0: int) -> List[List[int]]:
        ans = []
        for i in range(R):
            for j in range(C):
                ans.append([i,j])
        ans = sorted(ans, key=(lambda x : abs(x[0]-r0)+abs(x[1]-c0)))
        return ans

Java
class Solution {
    public int[][] allCellsDistOrder(int R, int C, int r0, int c0) {
        int[][] ans = new int[R*C][2];
        int k = 0;
        for(int i=0; i<R; ++i){
            for(int j=0; j<C; ++j){
                ans[k++] = new int[]{i,j};
            }
        }
        Arrays.sort(ans, new Comparator<int[]>(){
            public int compare(int[] a, int[] b){
                return Math.abs(a[0]-r0)+Math.abs(a[1]-c0) - Math.abs(b[0]-r0)-Math.abs(b[1]-c0);
            }
        });
        return ans;
    }
}