[LeetCode] 035: 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:
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].

[Solution]

/**
 * 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:
    // the start of the newInterval always later than the start of the last interval in intervals
    void insertInterval(vector<Interval> &intervals, Interval newInterval){
        // empty
        if(intervals.size() == 0){
            intervals.push_back(newInterval);
        }
        else{
            // the last of the intervals
            Interval interval = intervals[intervals.size()-1];
            
            // not intersected
            if(newInterval.start > interval.end){
                intervals.push_back(newInterval);
            }
            // intersected
            else if(newInterval.start <= interval.end){
                if(newInterval.end > interval.end){
                    intervals[intervals.size()-1].end = newInterval.end;
                }
            }
        }
    }
    // insert newInterval into intervals
    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        // result
        vector<Interval> res;
        bool inserted = false;  // whether the newInterval has been inserted
        for(int i = 0; i < intervals.size(); ++i){
            if(!inserted && newInterval.start <= intervals[i].start){
                inserted = true;
                insertInterval(res, newInterval);
            }
            insertInterval(res, intervals[i]);
        }
        
        // the intervals is empty or the newInterval is the last one
        if(!inserted){
            insertInterval(res, newInterval);
        }
        
        return res;
    }
};

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