notes/OJ notes/pages/Leetcode Merge-Intervals.md

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2022-09-01 14:14:08 +08:00
# 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;
}
};
```