leetcode : 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].

 

比较麻烦的是需要考虑newInerval 在第一个或者最后一个，或者插入在某个地方但是不需要合并的情况。

AC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> ret;
        if(intervals.size() == 0){
            ret.push_back(newInterval);
            return ret;
        }
        if(intervals[0].start > newInterval.end)                                           //假如newInterval在第一个
            ret.push_back(newInterval);
        int i = 0;
        while(i < intervals.size() && intervals[i].end < newInterval.start){
            ret.push_back(intervals[i++]);
        }
        if(i < intervals.size() && intervals[i].start > newInterval.end && ret[ret.size() - 1].end < newInterval.start)    //插入时不需要合并           
            ret.push_back(newInterval);
        for(; i < intervals.size(); ++i){
            if(intervals[i].start > newInterval.end)
                ret.push_back(intervals[i]);
            else{
                while(i < intervals.size() && intervals[i].start <= newInterval.end){              //归并所有需要归并的元素然后作为一个newInerval插入
                    newInterval.start = min(newInterval.start, intervals[i].start);
                    newInterval.end = max(newInterval.end, intervals[i].end);
                    ++i;
                }
                ret.push_back(newInterval);
                --i;
            }
        }
        if(ret[ret.size() - 1].end < newInterval.start)　　　　　　　　　　　　　　//在最后一个时
            ret.push_back(newInterval);
        return ret;
    }
};