[LeetCode] Merge Intervals、Insert Interval：

Merge Intervals:

Given a collection of intervals, merge all overlapping intervals.

For example,

Given

[1,3],[2,6],[8,10],[15,18]

,

return

[1,6],[8,10],[15,18]

.

/**
* 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 Interval_cmp:binary_function<Interval,Interval,bool>
{
public:
bool operator()(const Interval& i1,const Interval& i2)
{
if ( i1.start==i2.start )
return i1.end<i2.end;
return i1.start<i2.start;
}
};

class Solution {
public:
vector<Interval> merge(vector<Interval> &inters) {
// Start typing your C/C++ solution below
// DO NOT write int main() function

int n =inters.size();
if (n<=1 )
return inters;
sort(inters.begin(),inters.end(),Interval_cmp());
int start=inters[0].start,end=inters[0].end;
vector<Interval> ret;
for(int i=1;i<n;i++)
{
if ( end < inters[i].start )
{
ret.push_back(Interval(start,end));
start=inters[i].start;
end=inters[i].end;
}
else
{
end=max(end,inters[i].end);
}
}
ret.push_back(Interval(start,end));
return ret;
}
};

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]

.
注意O(n)一趟遍历的算法当然是很好想的，但打动不了面试官，既然都给你排序好了，当然要利用这点，用二分查找，这样算法是O(logN)的。

class Interval_cmp:public binary_function<Interval,int,bool>
{
public:
virtual bool operator()(const Interval& ,const int&) =0;
};
class Interval_cmp_start:public Interval_cmp
{
public:
bool operator()(const Interval& i1, const int& st)
{
return i1.start<st;
}
};

class Interval_cmp_end:public Interval_cmp
{
public:
bool operator()(const Interval& i1, const int& end)
{
return i1.end<end;
}
};
class Solution {
public:
vector<Interval> insert(vector<Interval> &inters, Interval newInterval) {
// Start typing your C/C++ solution below
// DO NOT write int main() function

vector<Interval> ret;
Interval_cmp_start start_pred=Interval_cmp_start();
Interval_cmp_end end_pred=Interval_cmp_end();

int start=newInterval.start;
int end=newInterval.end;
int idxStart= findInsertPos(inters,start,start_pred);
int idxEnd = findInsertPos(inters,end,end_pred);

if ( idxStart>0 && start<=inters[idxStart-1].end)
{
start=inters[idxStart-1].start;
idxStart--;
}
if (idxEnd<inters.size() && end>=inters[idxEnd].start)
{
end=inters[idxEnd].end;
idxEnd++;
}

for(int i=0;i<idxStart;i++)
ret.push_back(inters[i]);
ret.push_back(Interval(start,end));
for(int i=idxEnd;i<inters.size();i++)
ret.push_back(inters[i]);
return ret;
}
int findInsertPos(vector<Interval>& inters,int k,Interval_cmp& pred)
{
int l=0,r=inters.size()-1;
while(l<=r)
{
int mid=(l+r)>>1;
if ( pred(inters[mid],k) )
l=mid+1;
else
r=mid-1;
}
return l;
}

};