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[0]); for(int i = 1; i < intervals.size(); i++){ if(result[result.size()-1].end >= intervals[i].end) //totally contain continue; if(result[result.size()-1].end >= intervals[i].start){//there's overlap result[result.size()-1].end = intervals[i].end; } else{ //there's no overlap result.push_back(intervals[i]); } } return result; } };