LeetCode 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].

按照上一题的方法，时间复杂度O(N*log(N))

• 将给定区间push_back到整体区间；
• 然后排序，然后按照上一题的方法过一遍就可以了。

/**
 * 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 cmp(const Interval &a, const Interval &b)
 {
     return a.start < b.start;
 }
class Solution {
public:
    vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
        intervals.push_back(newInterval);
        sort(intervals.begin(),intervals.end(), cmp);

        vector<Interval> ans;

        int b = intervals[0].start, e = intervals[0].end;
        for(int i=1; i<intervals.size(); ++ i)
        {
            if(intervals[i].start > e)
            {
                ans.push_back(Interval(b, e));
                b = intervals[i].start, e = intervals[i].end;
            }
            else
            {
                if(intervals[i].end > e)
                    e = intervals[i].end;
            }
        }
        ans.push_back(Interval(b, e));
        return ans;
    }
};

利用数组有序，时间复杂度O(n)

• 每次比较给定区间的起始位置和当前期间的起始位置
• 根据起始位置的大小，选取起始位置小的区间
• 判断两个区间是否有交叉。有的话，当前区间往后挪动，直到没有交叉。然后break掉
• 将剩下的区间加入答案中。

/**
 * 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) {
        
        if(intervals.size() == 0)
            return vector<Interval>{newInterval, };
        
        vector<Interval> ans;
        int b, e, i;
        bool used = false;
        for(i=0; i<intervals.size(); ++ i)
        {
            if(intervals[i].start <= newInterval.start)
            {
                b = intervals[i].start, e = intervals[i].end;
                if(newInterval.start <= e)
                {
                    used = true;
                    e = max(e, newInterval.end);
                    while(i+1 < intervals.size() && intervals[i+1].start <= e)
                        i ++;
                    e = max(e, intervals[i].end);
                    i ++;
                }
            }
            else
            {
                used = true;
                b = newInterval.start, e = newInterval.end;
                if(newInterval.end >= intervals[i].start)
                {
                    e = max(e, intervals[i].end);
                    while(i+1 < intervals.size() && intervals[i+1].start <= e)
                        i ++;
                    e = max(e, intervals[i].end);
                    i ++;
                }
            }
            ans.push_back(Interval(b, e));
            if(used)
                break;
        }
        if(used == false)
            ans.push_back(newInterval);
        else
            for(; i<intervals.size(); ++ i)
                ans.push_back(intervals[i]);
        return ans;
    }
};