## Merge Intervals

• Time:O(n\log n)
• Space:O(n)

## C++

``````class Solution {
public:
vector<vector<int>> merge(vector<vector<int>>& intervals) {
vector<vector<int>> ans;

sort(begin(intervals), end(intervals));

for (const auto& interval : intervals)
if (ans.empty() || ans.back()[1] < interval[0])
ans.push_back(interval);
else
ans.back()[1] = max(ans.back()[1], interval[1]);

return ans;
}
};
``````

## JAVA

``````class Solution {
public int[][] merge(int[][] intervals) {
List<int[]> ans = new ArrayList<>();

Arrays.sort(intervals, (a, b) -> (a[0] - b[0]));

for (int[] interval : intervals)
if (ans.isEmpty() || ans.get(ans.size() - 1)[1] < interval[0])
else
ans.get(ans.size() - 1)[1] = Math.max(ans.get(ans.size() - 1)[1], interval[1]);

return ans.toArray(new int[ans.size()][]);
}
}
``````

## Python

``````class Solution:
def merge(self, intervals: List[List[int]]) -> List[List[int]]:
ans = []

for interval in sorted(intervals):
if not ans or ans[-1][1] < interval[0]:
ans.append(interval)
else:
ans[-1][1] = max(ans[-1][1], interval[1])

return ans
``````