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InsertInterval.h
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InsertInterval.h
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/*
Author: Annie Kim, anniekim.pku@gmail.com : King, wangjingui@outlook.com
Date: Jun 7, 2013
Update: Dec 14, 2014
Problem: Insert Interval
Difficulty: Medium
Source: https://oj.leetcode.com/problems/insert-interval/
Notes:
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].
Solution: For example 2:
1. compare [1,2] with [4,9], then insert [1,2];
2. merge [3,5] with [4,9], get newInterval = [3,9];
3. merge [6,7] with [3,9], get newInterval = [3,9];
4. merge [8,10] with [3,9], get newInterval = [3,10];
5. compare [12,16] with [3,10], insert newInterval [3,10], then all the remaining intervals...
Solution 1 : Time O(N).
Solution 2 : Time O(Log(N)).
*/
/**
* 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_1(vector<Interval> &intervals, Interval newInterval) {
vector<Interval> res;
vector<Interval>::iterator it = intervals.begin();
bool inserted = false;
for (; it != intervals.end(); ++it)
{
if (inserted == true || it->end < newInterval.start) // non-overlaping
{
res.push_back(*it);
}
else if (newInterval.end < it->start)
{
res.push_back(newInterval);
res.push_back(*it);
inserted = true;
}
else
{
newInterval.start = min(it->start, newInterval.start);
newInterval.end = max(it->end, newInterval.end);
}
}
if (inserted == false)
res.push_back(newInterval);
return res;
}
vector<Interval> insert_2(vector<Interval> &intervals, Interval newInterval) {
vector<Interval> res;
int n = intervals.size();
int left = 0, right = n-1;
while(left<=right){
int mid = (left+right)/2;
if(intervals[mid].start>newInterval.start) right=mid-1;
else left = mid+1;
}
int idxStart = right;
left = 0; right = n-1;
while(left<=right){
int mid = (left+right)/2;
if(intervals[mid].end>=newInterval.end) right = mid -1;
else left = mid+1;
}
int idxEnd = left;
if (idxStart>=0 && newInterval.start<=intervals[idxStart].end)
{
newInterval.start=intervals[idxStart--].start;
}
if (idxEnd<intervals.size() && newInterval.end>=intervals[idxEnd].start)
{
newInterval.end=intervals[idxEnd++].end;
}
for(int i=0;i<=idxStart;i++)
res.push_back(intervals[i]);
res.push_back(newInterval);
for(int i=idxEnd;i<intervals.size();i++)
res.push_back(intervals[i]);
return res;
}
};