Merge Intervals -- leetcode

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].

/**
 * 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> merge(vector<Interval> &intervals) {
        if (intervals.size() <= 1)
            return intervals;

        sort(intervals.begin(), intervals.end(), [](const Interval &a, const Interval &b) {return a.start < b.start;});

        size_t j = 0;
        for (size_t i = j+1; i<intervals.size(); i++) {
            if (intervals[j].end >= intervals[i].start)
                intervals[j].end = max(intervals[j].end, intervals[i].end);
            else
                intervals[++j] = intervals[i];
        }

        intervals.erase(intervals.begin()+j+1, intervals.end());
        return intervals;
    }
};

该算法在leetcode上实行执行时间为20ms。

基本思路：

1.利用自定的比较函数进行排序，以达到按start进行升序。

2. 利用remove-duplicates-from-sorted-array的思路进行归并