LeetCode57 Insert Interval

题目：

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]. （Hard）

分析：

首先可以采用merge interval的方法，先把区间填进去，然后排序，最后再调用merge，复杂度O(NlogN)

代码：

 /**
  * 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 {
 private:
     static bool cmp (const Interval& I1, const Interval& I2) {
         return I1.start < I2.start;
     }
     vector<Interval> merge(vector<Interval>& intervals) {
         vector<Interval> result;
         if (intervals.size() == ) {
             return result;
         }
         sort(intervals.begin(), intervals.end(), cmp);
         int left = intervals[].start, right = intervals[].end;
         for (int i = ; i < intervals.size(); ++i) {
             if (intervals[i].start <= right) {
                 right = max(right,intervals[i].end);
             }
             else {
                 result.push_back(Interval(left,right));
                 left = intervals[i].start;
                 right = intervals[i].end;
             }
         }
         result.push_back(Interval(left,right));
         return result;
     }
 public:
     vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
         vector<Interval> result;
         intervals.push_back(newInterval);
         sort(intervals.begin(),intervals.end(),cmp);
         result = merge(intervals);
         return result;
     }
 };

当然还有O(N)的算法可以优化。

分三步来做：

第一步，找到左侧不跟newInterval相交的区间添加到结果中；

第二步，找到所有和newInterval相交的区间并找到其左边界和右边界，然后建立新的interval添加到结果中；

第三部，找到右侧不跟newInterval相交的区间添加到结果中。

注意很多细节在里面可能会犯错

代码：

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