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  • 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) {
            int intervalsSize = intervals.size();
            vector<Interval> res;
            if(intervalsSize==0){
                res.push_back(newInterval);
                return res;
            }
            bool isFindStart = false,isFindEnd = false;
            int start = newInterval.start,end = newInterval.end;
            int startIndex = 0,endIndex=0;
            for(int i =0;i<intervalsSize;i++){//查找合并区间的起始和结束位置
                if(!isFindStart){
                    if(start >= intervals[i].start && start<=intervals[i].end){
                        isFindStart = true;
                        start = intervals[i].start;
                        startIndex = i;
                    }else if(start<intervals[i].start){
                        startIndex = i;
                        isFindStart = true;
                    }
                }
             //   cout<<"end="<<end<<",intervals["<<i<<"].start="<<intervals[i].start<<",intervals["<<i<<"].end="<<intervals[i].end<<endl;
                if(!isFindEnd){
                    if(end >= intervals[i].start && end<=intervals[i].end ){
                        isFindEnd = true;
                        end = intervals[i].end;
                        endIndex = i;
                    }else if(end<intervals[i].start){
                        endIndex = i-1;
                        isFindEnd = true;
                    }   
                }
                if(isFindStart && isFindEnd){
                    break;
                }
            }
            if(!isFindStart) startIndex = intervalsSize;
            for(int j=0;j<startIndex;j++){
                res.push_back(intervals[j]);
            }
            res.push_back(Interval(start,end));
            if(!isFindEnd) endIndex = intervalsSize-1;
            for(int j=endIndex+1;j<intervalsSize;j++){
                res.push_back(intervals[j]);
            }
            
            return res;
        }
    };
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  • 原文地址:https://www.cnblogs.com/zengzy/p/5006005.html
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