leetcode 57.Insert Interval

题意：

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].

 

刚开始遇到这个问题，我看了下hint，以为要用到binary tree，但是感觉上应该是挺简单的。后来发现却是是挺简单的

主要份两种情况：

1.newintervals 与原来的intervals 有交集，那么就需要合并

2.newintervals与原来的intervals 无交集，那么就不需要合并，直接按照大小排列即可

然后直接开一个新的数组，用于返回值，这样利用空间节省时间。

 

/**
 * 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> ret;
        if(intervals.empty())
        {
            ret.push_back(newInterval);
            return ret;
        }
            
        int i = 0;
        while(i < intervals.size())
        {
            //no overlapping
            if(newInterval.end < intervals[i].start)
            {
                ret.push_back(newInterval);
                while(i < intervals.size())
                {
                    ret.push_back(intervals[i]);
                    i ++;
                }
                return ret;
            }
            else if(newInterval.start > intervals[i].end)
                ret.push_back(intervals[i]);
            //overlapping
            else
            {
                newInterval.start = min(newInterval.start, intervals[i].start);
                newInterval.end = max(newInterval.end, intervals[i].end);
            }
            i ++;
        }
        ret.push_back(newInterval);      
        return ret;
    }
};