题目描述

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.

样例

Example 1:

Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]
Example 2:

Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].

算法1

(sort) $O(nlogn)$

存入新的interval，sort数组然后进行合并

时间复杂度

sort花费O(nlogn)
O(n)

参考文献

C++ 代码

class Solution {
public:
    vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
        intervals.push_back(newInterval);
        sort(intervals.begin(), intervals.end());
        vector<vector<int>> res;
        res.push_back(intervals[0]);        //先压入一个interval
        for(int i=1;i<intervals.size();i++)
        {
            vector<int> out=res.back();
            if(intervals[i][0]<=out[1])  
            {
                int end=max(out[1],intervals[i][1]);
                res.pop_back();
                res.push_back({out[0], end});
            }
            else
            {
                res.push_back(intervals[i]) ;                              
            }
        }

        return res;

    }
};

算法2

(遍历) $O(n)$

遍历数组，比较新的interval是否应该插入，然后合并

时间复杂度

遍历一遍，O(n)
O(n)

参考文献

C++ 代码

class Solution {
public:
    vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
        if(intervals.size()==0) return {newInterval};
        vector<vector<int>> res;
        bool st=false;
        int n=intervals.size(), insertstart=newInterval[0], insertend=newInterval[1];
        if(insertstart<intervals[0][0])
        {
            res.push_back(newInterval);
            st=true;
        }
        else
            res.push_back(intervals[0]);        //先压入一个interval

        for(int i=0;i<intervals.size();i++)
        {
            if(!st)
            {
                if(insertstart<intervals[i][0])
                {
                    res.push_back(newInterval);
                    st=true;
                }
            }
            vector<int> out=res.back();            
            if(intervals[i][0]<=out[1] )  
            {
                int end=max(out[1],intervals[i][1]);
                res.pop_back();
                res.push_back({out[0], end});
            }
            else
            {
                res.push_back(intervals[i]) ;                              
            }
            out=res.back();
            if(!st)
            {
                if(newInterval[0]<=out[1])
                {
                    int end=max(out[1],newInterval[1]);
                    res.pop_back();
                    res.push_back({out[0], end});
                    st=true;
                }
                else if(newInterval[0]>out[1] && (newInterval[0]<=intervals[i][0] ))
                {
                    res.push_back(newInterval);
                    st=true;
                }
            }

        }
        if(!st)
            res.push_back(newInterval);
        return res;

    }
};