1.8 KiB
1.8 KiB
Leetcode Merge-Intervals
2022-09-01 14:04
Algorithms:
#algorithm #sort
Data structures:
#DS #array
Difficulty:
#coding_problem #difficulty-medium
Additional tags:
#leetcode
Revisions:
N/A
Links:
Problem
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.
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 <= 104
intervals[i].length == 2
0 <= starti <= endi <= 104
Thoughts
[!summary] This is a generic array problem.
Situations to consider:
- The intervals can be unordered
- The first interval
[0, 3] [0, 1]
are adjacent and overlapped.
To solve the situations, sort first, and use max
function to solve the 3rd solution.
Solution
class Solution {
public:
vector<vector<int>> merge(vector<vector<int>> &intervals) {
// sort first, so that data are continious
vector<vector<int>> ans;
sort(intervals.begin(), intervals.end());
for (auto interval : intervals) {
if (ans.empty() || ans.back()[1] < interval[0]) {
ans.push_back(interval);
} else {
ans.back()[1] = max(interval[1], ans.back()[1]);
}
}
return ans;
}
};