[LeetCode] Merge Intervals

 

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].

 

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思路：首先按照start对interval排序，然后查看前一个的end是否和后一个的start相交，若相交，将pre标记成invalid，合并pre和cur，直至最后。不知道为啥这个题目被标记成hard

 另外注意： 采用自定义的比较函数，必须申明为全局函数，不能是类的成员函数。

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
bool cmp(Interval i, Interval j)
{
    return i.start < j.start;
}

class Solution {
    public:

        vector<Interval> merge(vector<Interval> &intervals)
        {
            size_t n = intervals.size();
            vector<bool> valid(n, true);
            vector<Interval> res;

            sort(intervals.begin(), intervals.end(), cmp);

#if 0
            for(int i = 0; i < intervals.size(); i++)
            {
                cout << "[" << intervals[i].start << "," << intervals[i].end << "]\t" ;
            }
            cout << endl;
#endif


            for(int i = 1; i < n; i++)
            {   
                Interval & pre = intervals[i-1];
                Interval & cur = intervals[i];

                if(pre.end >= cur.start)
                {   
                    valid[i-1] = false;

                    cur.start = pre.start;
                    if(pre.end > cur.end)
                        cur.end = pre.end;
                }
            }

            for(int i = 0; i < n; i++)
            {
//                cout << valid[i] << endl;
                if(valid[i] == true)
                    res.push_back(intervals[i]);
            }

            return res;
        }
};