https://leetcode.com/problems/merge-intervals/description/
56. Merge Intervals
这个题比较有意思的应该就是sort的写法。
/**
* 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:
vector<Interval> merge(vector<Interval>& intervals) {
vector<Interval> res;
if(intervals.size() < 1) return res;
sort(intervals.begin(), intervals.end(), compare);
Interval candi = intervals[0];
for(int i = 1; i < intervals.size(); i++)
{
if(intervals[i].start > candi.end)
{
res.push_back(candi);
candi = intervals[i];
}
else
{
candi.end = max(candi.end, intervals[i].end);
}
}
res.push_back(candi);
return res;
}
static bool compare(Interval a, Interval b)
{
return a.start == b.start ? a.end < b.end : a.start < b.start;
}
};
https://leetcode.com/submissions/detail/58964086/
57. Insert Interval
这种interval题貌似没有什么trick
逻辑清楚就好,分三部分,前面,中间,后面
前面就是比newInt.begin小的部分,后面就是比newInt.end大的部分,中间是需要merge的。
merge的方式都是找到最小的start,和最大的end
/**
* 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) {
vector<Interval>res;
Interval candi = newInterval;
int i = 0;
for(; i < intervals.size(); i++)
{
if(intervals[i].end < newInterval.start)
{
res.push_back(intervals[i]);
}
else
{
break;
}
}
for(; i < intervals.size() && intervals[i].start <= newInterval.end; i++)
{
candi.start = min(candi.start, intervals[i].start);
candi.end = max(candi.end, intervals[i].end);
}
res.push_back(candi);
for(; i < intervals.size(); i++)
{
res.push_back(intervals[i]);
}
return res;
}
};