# LeetCode 56. Merge Intervals

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_.

**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] overlaps, 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 <= 10^4`
* `intervals[i].length == 2`
* `0 <= starti <= endi <= 10^4`

## Solution

[English Version in Youtube](https://youtu.be/TBw99kT-r6A)

[中文版解答Youtube Link](https://youtu.be/WKEZOVgYVjI)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV1fK4y1N77s/)

```
class Solution {
public:
    vector<vector<int>> merge(vector<vector<int>> &intervals) {
        auto comp = [](const vector<int>& m, const vector<int>& n) {
            return m[0] < n[0];
        };
        sort(intervals.begin(), intervals.end(), comp);
        
        vector<int> now = intervals[0];
        vector<vector<int>> result;
        for (int i = 1;i < intervals.size();i++) {
            const vector<int>& then = intervals[i];
            if (now[1] >= then[0]) {
                now[1] = max(now[1], then[1]);
            } else {
                result.push_back(now);
                now = then;
            }
        }
        result.push_back(now);
        return result;
    }
};
```
