leetcode 56. Merge Intervals 57. Insert Interval

56. 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].

方法: map存点位，开始点+1，结束点-1。遍历map，count记录当前在这个点的段的数量。

class Solution {
public:
    vector<Interval> merge(vector<Interval>& intervals)
    {
        map<int, int> mp;
        for (auto it : intervals)
        {
            mp[it.start] ++;
            mp[it.end] --;
        }
        int count = 0;
        vector<Interval> ret;
        for (auto it : mp)
        {
            if (count == 0 && it.second > 0) //从无到有，加入头
            {
                ret.push_back(Interval(it.first, 0));
            }
            else if (count > 0 && it.second + count == 0) //从有到无,这一段结束
            {
                ret.back().end = it.first;
            }
            count += it.second;
        }
        return ret;
    }
};

57. 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].

同上。

class Solution {
public:
    vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
        map<int, int> mp;
        for (auto it : intervals)
        {
            mp[it.start] ++;
            mp[it.end] --;
        }
        mp[newInterval.start] ++;
        mp[newInterval.end] --;
        int count = 0;
        intervals.clear();
        for (auto it : mp)
        {
            if (count == 0 && it.second > 0) //从无到有，加入头
            {
                intervals.push_back(Interval(it.first, 0));
            }
            else if (count > 0 && it.second + count == 0) //从有到无,这一段结束
            {
                intervals.back().end = it.first;
            }
            count += it.second;
        }
        return intervals;
    }
};