# 【LeetCode】 56. Merge Intervals

## Description
> Given a collection of intervals, merge all overlapping intervals.
> NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

> 給予一個區間的集合，合併所有重疊到的區間。
> 注意: 2019年4月15日更改輸入型別，請重置成預設的程式碼並使用新的方法撰寫。

## Example:

Example 1:

Input: [[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: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

## Solution

* 先讓區間集合按大小排序。
* 設`s`是第一個區間的開頭、`e`是第一個區間的結尾：
* 如果接下來的區間落在`s`和`e`之間，就將`e`擴大。
* 如果沒有落在這區間，就執行合併。

### Code

class Solution {
public:
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        vector<vector<int>> ans;
        sort(intervals.begin(), intervals.end());
        
        bool first = false;
        int s, e;
        for(int i = 0; i < intervals.size(); i++)
        {
            if(!first)
            {
                first = true;
                s = intervals[i][0];
                e = intervals[i][1];
                if(i == intervals.size() - 1)
                    ans.push_back(vector<int>{s, e});
            }
            else if(intervals[i][0] <= e)
            {
                e = max(e, intervals[i][1]);
                if(i == intervals.size() - 1)
                    ans.push_back(vector<int>{s, e});
            }
            else
            {
                ans.push_back(vector<int>{s, e});
                first = false;
                i--;
            }
        }
        return ans;
    }
};

###### tags: `LeetCode` `C++`