56. Merge Intervals 57. Insert Interval *HARD*

Posted 2020-06-21 ArgenBarbie

1. Merge

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].

/**
 * 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:
    static bool compare(Interval a, Interval b)
    {
        if(a.start != b.start)
            return a.start < b.start;
        return a.end < b.end;
    }
    vector<Interval> merge(vector<Interval>& intervals) {
        sort(intervals.begin(), intervals.end(), compare);
        vector<Interval> v;
        int n = intervals.size(), i;
        if(n<1)
            return v;
        for(i = 0; i < n; i++)
        {
            if(v.size() && intervals[i].start <= v[v.size()-1].end)
                v[v.size()-1].end = max(intervals[i].end, v[v.size()-1].end);
            else
                v.push_back(intervals[i]);
        }
        return v;
    }
};

 

2. Insert

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].

(1) 用上面的merge方法

/**
 * 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:
    static bool compare(Interval a, Interval b)
    {
        if(a.start != b.start)
            return a.start < b.start;
        return a.end < b.end;
    }
    vector<Interval> merge(vector<Interval>& intervals) {
        sort(intervals.begin(), intervals.end(), compare);
        vector<Interval> v;
        int n = intervals.size(), i;
        if(n<1)
            return v;
        for(i = 0; i < n; i++)
        {
            if(v.size() && intervals[i].start <= v[v.size()-1].end)
                v[v.size()-1].end = max(intervals[i].end, v[v.size()-1].end);
            else
                v.push_back(intervals[i]);
        }
        return v;
    }
    vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
        intervals.push_back(newInterval);
        return merge(intervals);
    }
};

 

(2)

/**
 * 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) {
        int n = intervals.size(), l, i;
        vector<Interval> v;
        for(i = 0; i < n; i++)
        {
            if(newInterval.start > intervals[i].end)
                v.push_back(intervals[i]);
            else
                break;
        }
        if(i >= n)
        {
            v.push_back(newInterval);
            return v;
        }
        l = v.size();
        v.push_back(intervals[i]);
        if(newInterval.start < v[l].start)
            v[l].start = newInterval.start;
        while(i < n && newInterval.end > intervals[i].end)
            i++;
        if(i<n && newInterval.end >= intervals[i].start)
            v[l].end = intervals[i++].end;
        else
            v[l].end = newInterval.end;
        while(i<n)
            v.push_back(intervals[i++]);
        return v;
    }
};