• Insert Interval


    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].

    /**
     * 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> v;
            
            int l=-1;
            while(l+1<intervals.size() && intervals[l+1].end<newInterval.start) l=l+1;
            int r=intervals.size();
            while(r-1>=0 && intervals[r-1].start>newInterval.end) r=r-1;
            
            for(int i=0;i<=l;i++)
                v.push_back(intervals[i]);
            //
            l++;r--;
            if(l<intervals.size() && intervals[l].start<newInterval.start) newInterval.start=intervals[l].start;
            if(r>=0 && intervals[r].end>newInterval.end) newInterval.end=intervals[r].end;
            v.push_back(newInterval);
            //
            r++;
            for(int i=r;i<intervals.size();i++)
                v.push_back(intervals[i]);
            return v;
        }
    }; 
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  • 原文地址:https://www.cnblogs.com/erictanghu/p/3759421.html
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