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:

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].

解1:正常解法，时间复杂度O(n)

class Solution {
public:
    vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
        vector<vector<int>> res;
        int n = intervals.size();
        int i = 0;
        
        while(i < n && intervals[i][1] < newInterval[0])
        {
            res.push_back(intervals[i]);
            ++i;
        }
           
        while(i < n && intervals[i][0] <= newInterval[1])
        {
            newInterval[0] = min(newInterval[0], intervals[i][0]);
            newInterval[1] = max(newInterval[1], intervals[i][1]);
            ++i;
        }
        
        res.push_back(newInterval);
        while(i < n)res.push_back(intervals[i++]);
        return res;
    }
};

 

解2:利用STL函数equal_range()进行二分查找

class Solution {
public:
    vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
        auto compare = [](vector<int> a, vector<int> b) {return a[1] < b[0];};
        auto itRange = equal_range(intervals.begin(),intervals.end(),newInterval,compare);
        auto it1 = itRange.first, it2 = itRange.second;
        if (it1 == it2)
            intervals.insert(it1,newInterval);
        else
        {
            it2--;
            (*it2)[0] = min(newInterval[0],(*it1)[0]);
            (*it2)[1] = max(newInterval[1],(*it2)[1]);
            intervals.erase(it1,it2);
        }
        return intervals;
    }
};