【leetcode】Insert Interval

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 n=intervals.size();
        vector<Interval> result;
       
        if(n==0)
        {
            result.push_back(newInterval);
            return result;
        }
       
        int start=newInterval.start;
        int end=newInterval.end;
 
       
        bool isfindStart=false;
        bool isfindEnd=false;
        bool addCur=true;
       
        for(int i=0;i<n;i++)
        {
            addCur=true;
           
            if(!isfindStart)
            {
                //和当前区间没有交叉,且在start的前面
                if(start>intervals[i].end)
                {
                    //需要添加当前元素
                    addCur=true;
                    //当前元素为最后一个元素时，则认为start找到了
                    if(i==n-1)  isfindStart=true;
                }
                else
                {
                    //和当前区间有交叉 或者start小于当前区间
                   
                    start=min(start,intervals[i].start);
                    //后续就不用再找start了
                    isfindStart=true;
                   
                    //如果start和end没有和其他Interval有交叉
                    if(end<intervals[i].start) addCur=true;
                    else addCur=false;
                }
            }
           
           
            if(!isfindEnd)
            {
                //需要继续找end
                if(end>intervals[i].end)
                {
                   
                    if(start<=intervals[i].end) addCur=false;
 
                    if(i==n-1)
                    {
                        isfindEnd=true;
                       
                        if(addCur) result.push_back(intervals[i]);
                        result.push_back(Interval(start,end));
                       
                        continue;
                    }
                }
                else
                {
                    //end小于当前区间，此时找到了end
                    end=end>=intervals[i].start?intervals[i].end:end;
 
                    //找到了end
                    isfindEnd=true;
                    if(end>=intervals[i].start) addCur=false;
                   
                    result.push_back(Interval(start,end));
                    if(addCur) result.push_back(intervals[i]);
                   
                    continue;
                }
            }
           
            if(addCur) result.push_back(intervals[i]);
        }
        return result;
    }
};