题目:
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:
Input: intervals = [[1,3],[6,9]], newInterval = [2,5] Output: [[1,5],[6,9]]
Example 2:
Input: intervals =[[1,2],[3,5],[6,7],[8,10],[12,16]]
, newInterval =[4,8]
Output: [[1,2],[3,10],[12,16]] Explanation: Because the new interval[4,8]
overlaps with[3,5],[6,7],[8,10]
.
NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
代码:
/**
* 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> merged;
if(intervals.size() < 1)
return merged;
merged.push_back(intervals[0]);
int idx = 0;
for(int i = 1; i<intervals.size(); i++)
{
if(intervals[i].start >= merged[idx].start && intervals[i].start <= merged[idx].end)
{
merged[idx].start = min(merged[idx].start, intervals[i].start);
merged[idx].end = max(merged[idx].end, intervals[i].end);
}
else
{
merged.push_back(intervals[i]);
idx++;
}
}
return merged;
}
vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
if(intervals.size() < 1)
intervals.push_back(newInterval);
else
{
int insertFlag = 1;
if(newInterval.start < intervals[0].start || (newInterval.start == intervals[0].start && newInterval.end < intervals[0].end))
{
intervals.insert(intervals.begin(), newInterval);
insertFlag = 0;
}
for(int i = 1; i<intervals.size() && insertFlag; i++)
{
if(newInterval.start < intervals[i].start || (newInterval.start == intervals[i].start && newInterval.end <= intervals[i].end))
{
intervals.insert(intervals.begin()+i, newInterval);
insertFlag = 0;
}
}
if(insertFlag)
{
intervals.push_back(newInterval);
}
}
vector<Interval> mergeIntervals = merge(intervals);
return mergeIntervals;
}
};