## 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].

### Analysis:

In other words, the questions gives a new interval, the task is to insert this new interval into an ordered non-overlapping intervals. Consider the merge case.
Idea to solve this problem is quite straight forward:
1. Insert the new interval according to the start value.
2. Scan the whole intervals, merge two intervals if necessary.

### Code(C++):

/**
* 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:

bool myfunc(Interval a, Interval b) {
return( a.start<b.start);
}
vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<Interval> res;
vector<Interval>::iterator it;
for (it = intervals.begin();it!=intervals.end();it++){
if (newInterval.start<(*it).start){
intervals.insert(it,newInterval);
break;
}
}
if (it==intervals.end()){intervals.insert(it,newInterval);}

if (intervals.empty()) {return res;}

res.push_back(*intervals.begin());
for (it = intervals.begin()+1;it!=intervals.end();it++){
if ((*it).start>res.back().end){res.push_back(*it);}
else{ res.back().end = max(res.back().end,(*it).end);}
}
return res;
}
};


### Code(Python):

# Definition for an interval.
# class Interval:
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e

class Solution:
# @param intervals, a list of Intervals
# @param newInterval, a Interval
# @return a list of Interval
def insert(self, intervals, newInterval):
res = []
if len(intervals) == 0:
res.append(newInterval)
return res
# insert the new interval
intervals.append(newInterval)
# sort list according to the start value
intervals.sort(key=lambda x:x.start)

res.append(intervals[0])
#scan the list
for i in xrange(1,len(intervals)):
cur = intervals[i]
pre = res[-1]
#check if current interval intersects with previous one
if cur.start <= pre.end:
res[-1].end = max(pre.end, cur.end) #merge
else:
res.append(cur) #insert

return res



1. This comment has been removed by the author.

1. I have another solution:

class Solution:

def insert(self, intervals, newInterval):
if len(intervals) == 0:
return [newInterval]

v1 = self.search(intervals, newInterval.start, 0)
v2 = self.search(intervals, newInterval.end, 1)

l = []
if( v1 > 1):
for i in range(v1>>1):
l.append(intervals[i])

if v1 & 1 == 1:
newInterval.start = min(intervals[v1>>1].start, newInterval.start)
if v2 & 1 == 1:
newInterval.end = max(intervals[(v2)>>1].end, newInterval.end)

l.append(newInterval)

if v2 < (len(intervals)<<1 ) :
for i in range((v2+1)>>1, len(intervals)):
l.append(intervals[i])

return l

def search(self, intervals, index, isStart):
start = 0
end = len(intervals)-1
while(start <= end):
middle = ((end-start)>>1) + start
if intervals[middle].start <= index and intervals[middle].end >= index:
return (middle << 1)+1
if intervals[middle].start > index:
end = middle - 1
if intervals[middle].end < index:
start = middle + 1
return start << 1

url: http://traceformula.blogspot.com/2015/06/leetcode-insert-interval.html