1. Merge
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].
/**
* Definition for an interval.
* struct Interval {
* int start;
* int end;
* Interval() : start(0), end(0) {}
* Interval(int s, int e) : start(s), end(e) {}
* };
*/
class Solution {
public:
static bool compare(Interval a, Interval b)
{
if(a.start != b.start)
return a.start < b.start;
return a.end < b.end;
}
vector<Interval> merge(vector<Interval>& intervals) {
sort(intervals.begin(), intervals.end(), compare);
vector<Interval> v;
int n = intervals.size(), i;
if(n<)
return v;
for(i = ; i < n; i++)
{
if(v.size() && intervals[i].start <= v[v.size()-].end)
v[v.size()-].end = max(intervals[i].end, v[v.size()-].end);
else
v.push_back(intervals[i]);
}
return v;
}
};
2. Insert
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].
(1) 用上面的merge方法
/**
* Definition for an interval.
* struct Interval {
* int start;
* int end;
* Interval() : start(0), end(0) {}
* Interval(int s, int e) : start(s), end(e) {}
* };
*/
class Solution {
public:
static bool compare(Interval a, Interval b)
{
if(a.start != b.start)
return a.start < b.start;
return a.end < b.end;
}
vector<Interval> merge(vector<Interval>& intervals) {
sort(intervals.begin(), intervals.end(), compare);
vector<Interval> v;
int n = intervals.size(), i;
if(n<)
return v;
for(i = ; i < n; i++)
{
if(v.size() && intervals[i].start <= v[v.size()-].end)
v[v.size()-].end = max(intervals[i].end, v[v.size()-].end);
else
v.push_back(intervals[i]);
}
return v;
}
vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
intervals.push_back(newInterval);
return merge(intervals);
}
};
(2)
/**
* Definition for an interval.
* struct Interval {
* int start;
* int end;
* Interval() : start(0), end(0) {}
* Interval(int s, int e) : start(s), end(e) {}
* };
*/
class Solution {
public:
vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
int n = intervals.size(), l, i;
vector<Interval> v;
for(i = ; i < n; i++)
{
if(newInterval.start > intervals[i].end)
v.push_back(intervals[i]);
else
break;
}
if(i >= n)
{
v.push_back(newInterval);
return v;
}
l = v.size();
v.push_back(intervals[i]);
if(newInterval.start < v[l].start)
v[l].start = newInterval.start;
while(i < n && newInterval.end > intervals[i].end)
i++;
if(i<n && newInterval.end >= intervals[i].start)
v[l].end = intervals[i++].end;
else
v[l].end = newInterval.end;
while(i<n)
v.push_back(intervals[i++]);
return v;
}
};