57. Insert Interval ， STL的使用

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].

Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

/**
 * 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> insert(vector<Interval>& intervals, Interval newInterval) {
        auto compare = [] (const Interval &intv1, const Interval &intv2)
                          { return intv1.end < intv2.start; };//!!!!!!!!!注意了 不是return intv1.start < intv2.start; 
/*例如，A{1， 3} B{2,5}
根据自定义的比较函数有：A.end > B.start 为false 
                        B.end > A.start 为false
                    故：A = B;
              其意义是：如果A B重叠，那么A = B
所以，根据lower_bounder 二分查找后，如果找到的最左位置为index，返回index*/

        auto range = equal_range(intervals.begin(), intervals.end(), newInterval, compare);
        auto itr1 = range.first, itr2 = range.second;
        if (itr1 == itr2) {
            intervals.insert(itr1, newInterval);
        } else {
            itr2--;
            itr2->start = min(newInterval.start, itr1->start);
            itr2->end = max(newInterval.end, itr2->end);
            intervals.erase(itr1, itr2);//注意了！！！ 这里erase的是[itr1, itr2) 切记
        }
        return intervals;
    }
};