Leetcode 57. Insert Interval

Problem:

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].

Solution:

　　经典问题，两种做法。第一种，根据intervals的开始时间将newInterval插入到合适的位置，然后对intervals进行合并。第二种，找到完全在newInterval左侧的interval放到数组left中，找到完全在newInterva右侧的interval放到数组right中，对于有重合的interval向两头拓展start和end，最后将left+{start，end}+right合并为结果数组

Code:

 

/**
 * 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) {
        auto iter = intervals.begin();
        while(iter != intervals.end() && iter->start < newInterval.start)
            iter++;
        intervals.insert(iter,newInterval);
        vector<Interval> result;
        for(int i = 0;i != intervals.size();++i){
            if(result.empty() || intervals[i].start > result.back().end)
                result.push_back(intervals[i]);
            else{
                result.back().end = max(result.back().end,intervals[i].end);
            }
        }
        return result;
    }
};

 

/**
 * 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) {
        vector<Interval> left,right;
        vector<Interval> result;
        int start = newInterval.start;
        int end = newInterval.end;
        for(int i = 0;i != intervals.size();++i){
            if(end < intervals[i].start)
                right.push_back(intervals[i]);
            else if(start > intervals[i].end)
                left.push_back(intervals[i]);
            else{
                start = min(start,intervals[i].start);
                end = max(end,intervals[i].end);
            }
        }
        for(int i = 0;i != left.size();++i)
            result.push_back(left[i]);
        Interval interval(start,end);
        result.push_back(interval);
        for(int i = 0;i != right.size();++i)
            result.push_back(right[i]);
        return result;
    }
};