LeetCode(56) Insert Intervals

题目如下：

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].

分析如下：

由于输入的一堆intervals已经是排好序的了，所以接下来可以用二分查找找到new interval应该插入的合适的位置，这样新形成的intervals会比原来的intervals的数量多一。接下来对新的intervals进行merge就可以了, merge参考上面一道题目 Merge Intervals.

我的代码：

//72ms
/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
 //[[3,1],[5,6]], [1,1]
 //[[3,1],[5,6]], [7,8]
struct MyInterval{
    bool operator()(Interval interval1, Interval interval2) {
        return (interval1.start < interval2.start);
    }
} my_interval;

class Solution {
public:
    // void print_interval_vector(vector<Interval> interval_vec) {
    //     for (int i = 0; i < interval_vec.size(); ++i) {
    //         std::cout<<"["<<interval_vec[i].start<<", "<<interval_vec[i].end<<"]"<<", ";
    //     }
    //     std::cout<<std::endl;
    // }
    
    // vector<Interval> make_input(int input[][2], int len) {
    //     vector<Interval> res;
    //     for (int i = 0; i < len; ++i) {
    //         Interval a;
    //         a.start = input[i][0];
    //         a.end = input[i][1];
    //         res.push_back(a);
    //         std::cout<<"a.start = "<<a.start<<", a.end="<<a.end<<std::endl;
    //     }
    //     return res;
    // }

    vector<Interval> merge(vector<Interval> &intervals) {
        vector<Interval> final;
        //std::sort(intervals.begin(), intervals.end(), my_interval); //unnecessary to sort
        final.push_back(intervals[0]);
        for (int i = 1; i < intervals.size(); ++i) {
            if (intervals[i].start <= final.back().end) {
                final.back().end = intervals[i].end > final.back().end ? intervals[i].end : final.back().end;
            } else {
                final.push_back(intervals[i]);
            }
        }
        return final;
    }
    
    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
        vector<Interval> result;
        if (intervals.size() == 0) {
            result.push_back(newInterval);
            return vector<Interval> result;
        }
        // first, insert
        int start_index = 0, end_index = intervals.size(), mid_index = 0, insert_index = -1;
        //std::sort(intervals.begin(), intervals.end(), my_interval); //unnecessary to sort
        while (start_index <= end_index) {
            mid_index = start_index + (end_index - start_index) / 2;
            if (intervals[mid_index].start == newInterval.start) {
                insert_index = mid_index;
                break;
            } else if (intervals[mid_index].start > newInterval.start) {
                end_index = mid_index - 1;
            } else {
                start_index = mid_index + 1;//this can be concluded from the example above
            }
        }
        if (insert_index == -1)
            insert_index = start_index;
        if (insert_index > intervals.size()) insert_index = intervals.size();
        if (insert_index < 0) insert_index = 0;
            intervals.insert(intervals.begin() + insert_index, newInterval);
        //second, merge
        return  merge(intervals);
    }
};