57. Insert Interval

problem description

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.

algorithm thought

和上一题一样，这里更简单的是，已经直接排好序了。所以，我们只需要找出中间可能有交叉的interval。然后合并。前后没有交叉的interval，就不变，直接插入。

code

/**
 * 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;
        int index=0;
        while(index<intervals.size()&&intervals[index].end<newInterval.start)
            res.push_back(intervals[index++]);
        while(index<intervals.size()&&intervals[index].start<=newInterval.end){
            newInterval.start=min(newInterval.start,intervals[index].start);
            newInterval.end=max(newInterval.end,intervals[index].end);
            index++;
        }
        res.push_back(newInterval);
        while(index<intervals.size()&&index!=intervals.size())
            res.push_back(intervals[index++]);
        return res;
    }
};

algorithm analysis

这里不需要排序，时间复杂度O(n)