Insert Interval

Q:

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].

 

A: 复杂度O(n).

Interval数组互不相交且有序，因此这些Intervals从小到大排序。遍历所有的Interval，首先将所有排在newInterval之前，拷贝到结果，接着merge与NewInterval相交的Interval(Intervals[i].start<=NewInterval.end)，merge后更新NewInteral的start、end，将新的NewInterval放到结果中，最后将所有排在NewInterval后面的放到结果中。

    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<Interval> res;
        
        int n = intervals.size();
        int i = 0;
        
        while(i<n&&intervals[i].end<newInterval.start)
            res.push_back(intervals[i++]);
            
        while(i<n&&intervals[i].start<=newInterval.end)
        {
            newInterval.start = min(newInterval.start,intervals[i].start);
            newInterval.end = max(newInterval.end,intervals[i].end);
            ++i;
        }
        
        res.push_back(newInterval);
        
        while(i<n)
            res.push_back(intervals[i++]);
        return res;
        
    }