LeetCode56. Merge Intervals（C++）

Given a collection of intervals, merge all overlapping intervals.

Example 1:

Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].

Example 2:

Input: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

解题思路：贪心法，按照每个区间的左端点进行排序后，两个区间只要有交集就进行合并，则对数个待合并的区间来说，合并后的左端点为第一个区间的左端点，右端点为这些区间的最远右端点。所以在遍历过程中，只要记录更新最远右端点，当下一个区间的左端点不包括在该最远右端点中，则进行下一个合并区间。

/**
 * 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 merge(vector& intervals) {
        sort(intervals.begin(), intervals.end(), [](Interval a, Interval b){return a.start < b.start;});
        vectorresult;
        int i = 0;
        while(i < intervals.size()){
            int low = intervals[i].start,high = intervals[i].end;
            while(++ i < intervals.size() && intervals[i].start <= high)
                high = max(high,intervals[i].end);
            Interval temp;
            temp.start = low;temp.end=high;
            result.push_back(temp);
        }
        return result;
    }
   
};