LeetCode题解：Merge/Insert Interval

Merge Intervals

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].

Insert Interval

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].

思路：

第一题的思路是先排序，然后依次向下比较归并。

第二题可以先插入再用第一题的归并方法。但是这里采用计算interval的覆盖位置的方法，相对复杂一些。

题解：

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */

bool operator< (const Interval& i1, const Interval& i2)
{
    return i1.start < i2.start;
}

class Solution {
public:
    bool try_merge_next (Interval& i1, const Interval& i2)
    {
        if (i1.end < i2.start)
            return false;   // not contacting
        else
        {
            i1.end = max (i1.end, i2.end);
            return true;
        }
    }
    
    vector<Interval> merge (vector<Interval>& intervals)
    {
        vector<Interval> ret;
        
        sort (begin (intervals), end (intervals));
    
        if (!intervals.empty())
        {
            ret.push_back (intervals.front());
            for (auto & i : intervals)
                if (!try_merge_next (ret.back(), i))
                    ret.push_back (i);
        }        
        return ret;
    }
};

/**
 * 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:
    bool is_prior (const Interval& i, int val)
    {
        return (i.start < val && i.end < val);
    }
    
    bool is_after (const Interval& i, int val)
    {
        return (i.start > val && i.end > val);
    }
    
    vector<Interval> insert (vector<Interval>& intervals, Interval ni)
    {
        vector<Interval> ret;
        
        if (intervals.empty())
        {
            ret.push_back (ni);
            return ret;
        }
        
        if (is_after (intervals.front(), ni.end))
        {
            ret.push_back (ni);
            copy (begin (intervals), end (intervals), back_inserter (ret));
            return ret;
        }
        
        if (is_prior (intervals.back(), ni.start))
        {
            copy (begin (intervals), end (intervals), back_inserter (ret));
            ret.push_back (ni);
            return ret;
        }
        
        // now we have to find the exact position
        auto iter = begin (intervals);
        auto last = end (intervals);
        
        while (is_prior (*iter, ni.start))
            ret.push_back (*iter++);
            
        ni.start = min (ni.start, iter->start);
        
        while (iter != last && !is_after (*iter, ni.end)) ++iter;
        
        ni.end = max (ni.end, (iter - 1)->end);
        
        ret.push_back (ni);
        copy (iter, last, back_inserter (ret));
        
        return ret;
    }
};