1203.Sort Items by Groups Respecting Dependencies
===
###### tags: `Hard` `DFS` `BFS` `Graph` `Topological Sort`
[1203. Sort Items by Groups Respecting Dependencies](https://leetcode.com/problems/sort-items-by-groups-respecting-dependencies/)
### 題目描述
There are `n` items each belonging to zero or one of `m` groups where `group[i]` is the group that the i-th item belongs to and it's equal to `-1` if the `i`-th item belongs to no group. The items and the groups are zero indexed. A group can have no item belonging to it.
Return a sorted list of the items such that:
* The items that belong to the same group are next to each other in the sorted list.
* There are some relations between these items where `beforeItems[i]` is a list containing all the items that should come before the `i`-th item in the sorted array (to the left of the `i`-th item).
Return any solution if there is more than one solution and return an **empty list** if there is no solution.
### 範例
**Example 1:**
![](https://assets.leetcode.com/uploads/2019/09/11/1359_ex1.png)
```
Input: n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3,6],[],[],[]]
Output: [6,3,4,1,5,2,0,7]
```
**Example 2:**
```
Input: n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3],[],[4],[]]
Output: []
Explanation: This is the same as example 1 except that 4 needs to be before 6 in the sorted list.
```
**Constraints**:
* 1 <= `m` <= `n` <= 3 * 10^4^
* `group.length` == `beforeItems.length` == `n`
* -1 <= `group[i]` <= `m` - 1
* 0 <= `beforeItems[i].length` <= `n` - 1
* 0 <= `beforeItems[i][j]` <= `n` - 1
* `i` != `beforeItems[i][j]`
* `beforeItems[i]` does not contain duplicates elements.
### 解答
### Reference
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