79 lines
1.8 KiB
Markdown
79 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;
|
|||
|
}
|
|||
|
};
|
|||
|
```
|