**source:**https://leetcode.com/problems/merge-intervals**C/C++**

**Solution to LeetCode**problem

**56**.

**Merge Intervals**.

## Problem

Given an array of `intervals`

where `intervals[i] = [start`

, merge all overlapping intervals, and return _{i}, end_{i}]*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 <= 10`

^{4}`intervals[i].length == 2`

`0 <= start`

_{i}<= end_{i}<= 10^{4}

## Solution

We need to sort the initial vector, so that every range will have a `start`

value _{i}`>=`

than the previous range.

- Once sorted, we push the first range to the result.
- We iterate through all the next elements.
- If
`start`

_{i}>= start_{i-1}- If
`end`

we have a new end for the last element in the results._{i}>= end_{i-1}

- If
- If not, we have a new range to add to the results.

- If

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class Solution {
public:
vector<vector<int>> merge(vector<vector<int>>& intervals) {
if (intervals.size() == 0) return { {} };
vector<vector<int>> result;
sort(intervals.begin(), intervals.end());
result.push_back(intervals[0]);
for (int i=1; i<intervals.size(); i+=1) {
if (intervals[i][0] >= result[result.size()-1][0] &&
intervals[i][0] <= result[result.size()-1][1]) {
if(intervals[i][1] >= result[result.size()-1][1])
result[result.size()-1][1] = intervals[i][1];
} else {
result.push_back(intervals[i]);
}
}
return result;
}
};