Given a collection of intervals, merge all overlapping intervals.
For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18].
注意点：
1。并不是像例子中给出的那样，前一个interval必定在后一个interval的前边。
所以我们先要对start进行排序。必须把start小的放在前面，然后按序递增，否则会出现这样的错误
Input:[[2,3],[4,5],[6,7],[8,9],[1,10]]
Output:[[2,3],[4,5],[6,7],[1,10]]
Expected:[[1,10]]
2。sort start需要自定义compare函数，注意升序的时候不能定义<=，否则会造成Time Limit Exceeded
原因是sort的compare函数必须满足strict weak ordering。时间复杂度O(nlogn), reference: http://www.cplusplus.com/reference/list/list/sort/
3。两个字串长度和 =各子串长度相加的时候，如下例，并没有overlap。仅当[1,4],[4,6]才有overlap。
[[1,4],[5,6]]
Output:[[1,6]]
Expected:[[1,4],[5,6]]
/**
* 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:
static bool compare(Interval v1, Interval v2)
{
if(v1.start < v2.start)
return true;
else if(v1.start > v2.start)
return false;
else
return v1.end < v2.end;
} vector<Interval> merge(vector<Interval>& intervals) {
vector<Interval> result;
if(intervals.empty()) return result; sort(intervals.begin(),intervals.end(),compare);
result.push_back(intervals[]);
for(int i = ; i < intervals.size(); i++){
if(result[result.size()-].end >= intervals[i].end) //totally contain
continue; if(result[result.size()-].end >= intervals[i].start){//there's overlap
result[result.size()-].end = intervals[i].end;
}
else{ //there's no overlap
result.push_back(intervals[i]);
}
} return result;
}
};