Merge Intervals – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
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] 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
C++ Merge Intervals LeetCode Solution
vector<Interval> merge(vector<Interval>& ins) {
if (ins.empty()) return vector<Interval>{};
vector<Interval> res;
sort(ins.begin(), ins.end(), [](Interval a, Interval b){return a.start < b.start;});
res.push_back(ins[0]);
for (int i = 1; i < ins.size(); i++) {
if (res.back().end < ins[i].start) res.push_back(ins[i]);
else
res.back().end = max(res.back().end, ins[i].end);
}
return res;
}
Java Merge Intervals LeetCode Solution
class Solution {
public int[][] merge(int[][] intervals) {
if (intervals.length <= 1)
return intervals;
// Sort by ascending starting point
Arrays.sort(intervals, (i1, i2) -> Integer.compare(i1[0], i2[0]));
List<int[]> result = new ArrayList<>();
int[] newInterval = intervals[0];
result.add(newInterval);
for (int[] interval : intervals) {
if (interval[0] <= newInterval[1]) // Overlapping intervals, move the end if needed
newInterval[1] = Math.max(newInterval[1], interval[1]);
else { // Disjoint intervals, add the new interval to the list
newInterval = interval;
result.add(newInterval);
}
}
return result.toArray(new int[result.size()][]);
}
}
Python 3 Merge Intervals LeetCode Solution
def merge(self, intervals):
out = []
for i in sorted(intervals, key=lambda i: i.start):
if out and i.start <= out[-1].end:
out[-1].end = max(out[-1].end, i.end)
else:
out += i,
return out
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