合并区间-LintCode

给出若干闭合区间，合并所有重叠的部分。
样例
给出的区间列表 => 合并后的区间列表：
[
[
[1, 3], [1, 6],
[2, 6], => [8, 10],
[8, 10], [15, 18]
[15, 18] ]
]
挑战
O(n log n) 的时间和 O(1) 的额外空间。

#ifndef C156_H#define C156_H#include<iostream>#include<vector>#include<algorithm>using namespace std;class Interval{public:    int start, end;    Interval(int start, int end)    {        this->start = start;        this->end = end;    }};bool comp(const Interval &a, const Interval &b){    return a.start < b.start;}class Solution {public:    /**    * @param intervals: interval list.    * @return: A new interval list.    */    vector<Interval> merge(vector<Interval> &intervals) {        // write your code here        if (intervals.empty())            return intervals;        sort(intervals.begin(), intervals.end(),comp);        int len = intervals.size();        int i = 0;        while (i < len-1)        {            if (intervals[i].start<=intervals[i+1].start&&intervals[i].end>=intervals[i+1].end)            {                intervals[i + 1] = intervals[i];                intervals.erase(intervals.begin() + i);                len--;            }            else if (intervals[i].end >= intervals[i + 1].start)            {                intervals[i + 1].start = intervals[i].start;                intervals.erase(intervals.begin() + i);                len--;            }            else            {                i++;            }        }        return intervals;    }};#endif