92 lines
1.8 KiB
Markdown
92 lines
1.8 KiB
Markdown
# Leetcode Merge-Intervals
|
||
|
||
#### 2022-09-01 14:04
|
||
|
||
> ##### Algorithms:
|
||
>
|
||
> #algorithm #sort
|
||
>
|
||
> ##### Data structures:
|
||
>
|
||
> #DS #array
|
||
>
|
||
> ##### Difficulty:
|
||
>
|
||
> #coding_problem #difficulty-medium
|
||
>
|
||
> ##### Additional tags:
|
||
>
|
||
> #leetcode
|
||
>
|
||
> ##### Revisions:
|
||
>
|
||
> N/A
|
||
|
||
##### Links:
|
||
|
||
- [Link to problem](https://leetcode.com/problems/merge-intervals/)
|
||
- [Solution ans Explanation](https://leetcode.com/problems/merge-intervals/solution/)
|
||
|
||
---
|
||
|
||
### Problem
|
||
|
||
Given an array of `intervals` where `intervals[i] = [starti, endi]`, merge all overlapping intervals, and return _an array of the non-overlapping intervals that cover all the intervals in the input_.
|
||
|
||
#### Examples
|
||
|
||
**Example 1:**
|
||
|
||
**Input:** intervals = [[1,3],[2,6],[8,10],[15,18]]
|
||
**Output:** [[1,6],[8,10],[15,18]]
|
||
**Explanation:** Since intervals [1,3] and [2,6] overlap, merge them into [1,6].
|
||
|
||
**Example 2:**
|
||
|
||
**Input:** intervals = [[1,4],[4,5]]
|
||
**Output:** [[1,5]]
|
||
**Explanation:** Intervals [1,4] and [4,5] are considered overlapping.
|
||
|
||
#### Constraints
|
||
|
||
- `1 <= intervals.length <= 104`
|
||
- `intervals[i].length == 2`
|
||
- `0 <= starti <= endi <= 104`
|
||
|
||
### Thoughts
|
||
|
||
> [!summary]
|
||
> This is a generic array problem.
|
||
|
||
#### Situations to consider:
|
||
|
||
- The intervals can be unordered
|
||
- The first interval
|
||
- `[0, 3] [0, 1]` are adjacent and overlapped.
|
||
|
||
To solve the situations, sort first, and use `max` function to solve the 3rd solution.
|
||
|
||
### Solution
|
||
|
||
```cpp
|
||
class Solution {
|
||
public:
|
||
vector<vector<int>> merge(vector<vector<int>> &intervals) {
|
||
// sort first, so that data are continious
|
||
vector<vector<int>> ans;
|
||
|
||
sort(intervals.begin(), intervals.end());
|
||
|
||
for (auto interval : intervals) {
|
||
if (ans.empty() || ans.back()[1] < interval[0]) {
|
||
ans.push_back(interval);
|
||
} else {
|
||
ans.back()[1] = max(interval[1], ans.back()[1]);
|
||
}
|
||
}
|
||
|
||
return ans;
|
||
}
|
||
};
|
||
```
|