每日算法之四十：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].

这个跟之前的合并类似，最好也是逐一考虑即可，记录好newInterval即可，记得最后要插入一次，函数有两个出口，两个出口都要处理好修改后的newInterval。

/**
 * 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
    //Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].
      int size = intervals.size();
      vector<Interval> res;
      for(int i = 0; i < size; i++)
      {
          if(intervals[i].end < newInterval.start )
                res.push_back(intervals[i]);
          else if(intervals[i].start > newInterval.end)
          {
                res.push_back(newInterval);
                res.insert(res.end(),intervals.begin() + i, intervals.end() );
                return res;
          }
          else
          {
              newInterval.start = min(newInterval.start, intervals[i].start);
              newInterval.end   = max(newInterval.end, intervals[i].end);
          }
      }
      res.push_back(newInterval);
      return res;
    }
 };
    

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