LeetCode(56) Insert Intervals

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本文介绍了一种在已排序的非重叠区间中插入新区间并进行必要合并的方法。通过二分查找确定新区间的位置，并对合并后的区间进行进一步处理。

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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);
    }
};