# Leetcode日记—1030. Matrix cells in Distance order

## 题目如下：

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.)

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]].

Note:

1 <= R <= 100
1 <= C <= 100
0 <= r0 < R
0 <= c0 < C

## 思路：

### 方法一：

``````class Solution {
public int[][] allCellsDistOrder(int R, int C, int r0, int c0) {
int[][] res = new int[R*C];
//cur用于建立答案数组
int cur = 0;
for(int i=0;i<R;i++){
for(int j=0;j<C;j++){
res[cur] = i;
res[cur] = j;
cur+=1;
}
}
Arrays.sort(res,new Comparator<>(){
public int compare(int[] a,int[] b){
return (Math.abs(a - r0) + Math.abs(a - c0)) - (Math.abs(b - r0) + Math.abs(b - c0));
}
});
return res;
}
}``````

### 方法二：

``````class Solution {
public int[][] allCellsDistOrder(int R, int C, int r0, int c0) {
int maxDist = Math.max(r0, R - 1 - r0) + Math.max(c0, C - 1 - c0);
List<List<int[]>> bucket = new ArrayList<List<int[]>>();
for (int i = 0; i <= maxDist; i++) {
}

for (int i = 0; i < R; i++) {
for (int j = 0; j < C; j++) {
int d = dist(i, j, r0, c0);
}
}
int[][] ret = new int[R * C][];
int index = 0;
for (int i = 0; i <= maxDist; i++) {
for (int[] it : bucket.get(i)) {
ret[index++] = it;
}
}
return ret;
}

public int dist(int r1, int c1, int r2, int c2) {
return Math.abs(r1 - r2) + Math.abs(c1 - c2);
}
}