leetcode 57 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].

 

First, this problem isn't about segment tree.

 

Second, I wanna share a terrible code without using the function insert and considering the case [[3,5],[12,15]], [6,6]

/**
 * 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) {
    // Note: The Solution object is instantiated only once and is reused by each test case.
    vector<Interval>::iterator it = intervals.begin(), start, end;
    if (intervals.size() == 0) {
      intervals.clear();
      intervals.push_back(newInterval);
    }
    else if (newInterval.start <= intervals[0].start || newInterval.end >= intervals[intervals.size() - 1].end){
      newInterval.start = min(newInterval.start, intervals[0].start);
      newInterval.end = max(newInterval.end, intervals[intervals.size() - 1].end);
      intervals.clear();
      intervals.push_back(newInterval);
    }
    else {
      while (it != intervals.end()) {
        if (newInterval.start >= (*it).start && newInterval.start <= (*it).end)
          start = it;
        if (newInterval.end >= (*it).start && newInterval.end <= (*it).end) {
          end = it;
          break;
        }
        ++it;
      }
      int s1 = (*start).start;
      int s2 = (*end).end;
      
      it = start;
      while (it != end) {
        intervals.erase(it);
        ++it;
      }
      
      (*it).start = s1;
      (*it).end = s2;        
    }
    return intervals;
  }
};

When using insert, a concise code is like:

/**
 * 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) {
    // Note: The Solution object is instantiated only once and is reused by each test case.
    vector<Interval>::iterator it = intervals.begin();
    while (it != intervals.end()) {
      if (newInterval.start > it->end) 
        ++it;
      else if (newInterval.end < it->start) {
        intervals.insert(it, newInterval);
        return intervals;
      }
      else{
        newInterval.start = min(newInterval.start,it->start);
        newInterval.end = max(newInterval.end,it->end);
        it = intervals.erase(it);
      }
    }
    intervals.insert(intervals.end(), newInterval);
    return intervals;
  }
};

新new一个结果看起来更加干净一些：

def insert(self, intervals, newInterval):
    s, e = newInterval.start, newInterval.end
    left, right = [], []
    for i in intervals:
        if i.end < s:
            left += i,
        elif i.start > e:
            right += i,
        else:
            s = min(s, i.start)
            e = max(e, i.end)
    return left + [Interval(s, e)] + right