Leetcode:57. Insert Interval

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

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].

思路

• 首先，如果和newInterval区间不相交的区间，可以直接排除，也就是前面两个if，后面那个else则是重新修改newInterval区间用的

代码

/**
 * 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 len = intervals.size();
        if(len == 0){
            res.push_back(newInterval);
            return res;
        } 
        
        int i = 0;
        bool flag = false;
        while(i < len){
            Interval tmp;
            tmp.start = intervals[i].start;
            tmp.end = intervals[i].end;
           
           if(tmp.end < newInterval.start){
                res.push_back(tmp);
                ++i;
           }
           else if(tmp.start > newInterval.end){
                if(!flag){
                    res.push_back(newInterval);
                    flag = true;
                }
                res.push_back(tmp);
                ++i;
            }
            else{
                newInterval.start = min(newInterval.start, tmp.start);
                newInterval.end = max(newInterval.end, tmp.end);
                ++i;
            }
        }
        
        if(!flag)
            res.push_back(newInterval);
        
        return res;
    }
};