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].
class Solution {
public:
vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
vector<Interval> result;
for(int i=0;i<intervals.size();i++) {
if(intervals[i].end < newInterval.start) {
result.push_back(intervals[i]);
}
else if(intervals[i].start > newInterval.end) {
result.push_back(newInterval);
result.insert(result.end(),intervals.begin()+i,intervals.end());
return result;
}
else {
newInterval.start = min(newInterval.start, intervals[i].start);
newInterval.end = max(newInterval.end, intervals[i].end);
}
}
result.push_back(newInterval);
return result;
}
};原文：http://blog.csdn.net/starcuan/article/details/18853431