LeetCode: 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].

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<Interval> result;
        int nSize = intervals.size();
        if (nSize == 0)
        {
            result.push_back(newInterval);
            return result;
        }
        
        int idx = 0;
        Interval val;
        for (; idx < nSize; ++idx)
        {
            val = intervals[idx];
            if (newInterval.start > val.end)
            {
                 result.push_back(val);
            }
            // 无相交区域
            else if (newInterval.end < val.start)
            {
                result.push_back(newInterval);
                while(idx < nSize)
                    result.push_back(intervals[idx++]);
            }
            else
            {
                newInterval.start = newInterval.start < val.start ?
                                    newInterval.start : val.start;
                newInterval.end =  newInterval.end > val.end ?
                                    newInterval.end : val.end;
            }
        }
        
        if (result.size() == 0 || result.back().end < newInterval.start)
        {
            result.push_back(newInterval);
        }
        return result;
    }
};