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].

 

向已经有序的且没重合的区间集合中插入新区间，并合并由此引来的重合区间，输出。

/**
 * 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> ans;   //存放结果
        int i=0;
        while( i<intervals.size() && intervals[i].end < newInterval.start ) //如果旧区间的end小于新区间的start，那么直接放入结果中
            ans.push_back( intervals[i++] );
        if( i == intervals.size() ) {   //处理特定的情况，及新区间的start比旧区间最后一个区间的end还大
            ans.push_back(newInterval);
            return ans;
        }
        //不断合并重复区间，直到第一个旧区间的start大于新区间的end
        while( i<intervals.size() && intervals[i].start <= newInterval.end ) {
            newInterval.start = min(intervals[i].start, newInterval.start);
            newInterval.end = max(intervals[i].end, newInterval.end);
            ++i;
        }
        ans.push_back(newInterval);
        while( i<intervals.size() ) ans.push_back(intervals[i++]);  //将后面不需要合并的区间放入结果中
        return ans;
    }
};