Intervals structure
struct Interval {
int start;
int end;
Interval() : start(0), end(0) {}
Interval(int s, int e) : start(s), end(e) {}
};
Leetcode 56. Merge Intervals
Given a collection of intervals, merge all overlapping intervals.
For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18].
class Solution {
public:
vector<Interval> merge(vector<Interval>& intervals) {
vector<Interval> res;
if (intervals.empty()) return res;
sort(intervals.begin(), intervals.end(), [](Interval &i, Interval &j){return i.start<j.start;});
const int n = intervals.size();
for (int i=0; i<n; i++) {
if (i+1<n && intervals[i+1].start<=intervals[i].end) {
intervals[i+1].end = max(intervals[i].end, intervals[i+1].end);
intervals[i+1].start = min(intervals[i].start, intervals[i+1].start);
}
else
res.push_back(intervals[i]);
}
return res;
}
};
Leetcode 57. Insert Interval
Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).
You may assume that the intervals were initially sorted according to their start times.
Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].
Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].
This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].
Push back to the vector and then use the previous merge function.
class Solution {
public:
vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
vector<Interval> res;
intervals.push_back(newInterval);
sort(intervals.begin(), intervals.end(), [](Interval &i, Interval &j){return i.start<j.start;});
const int n = intervals.size();
for (int i=0; i<n; i++) {
if (i+1<n && intervals[i+1].start<=intervals[i].end) {
intervals[i+1].end = max(intervals[i].end, intervals[i+1].end);
intervals[i+1].start = min(intervals[i].start, intervals[i+1].start);
}
else
res.push_back(intervals[i]);
}
return res;
}
};