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Insert_Interval.cpp
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127 lines (105 loc) · 3.58 KB
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/*
Author: Timon Cui, [email protected]
Title: Insert Interval
Description:
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].
Difficulty rating:
Source:
http://www.leetcode.com/onlinejudge
Notes:
[ ] [ ] [ ] [ ] [ ]
( b ] ib
( e ] ie
Binary search for b and e in the list.
remove everything between b and e
insert new interval = min(b, begin after b), max(e, end after e)
*/
/**
* Definition for an interval.
* struct Interval {
* int start;
* int end;
* Interval() : start(0), end(0) {}
* Interval(int s, int e) : start(s), end(e) {}
* };
*/
#include "utils.hpp"
using namespace std;
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) {
int ib = lower_bound(intervals.begin(), intervals.end(), newInterval, compare) - intervals.begin();
int ie = lower_bound(intervals.begin(), intervals.end(), Interval(newInterval.end, 0), compare) - intervals.begin();
if (ib > 0 && newInterval.start <= intervals[ib - 1].end) {
newInterval.start = intervals[ib - 1].start;
ib --;
}
if (ie > 0) newInterval.end = max(newInterval.end, intervals[ie - 1].end);
if (ie < intervals.size() && newInterval.end >= intervals[ie].start) {
newInterval.end = intervals[ie].end;
ie ++;
}
vector<Interval> result(intervals.begin(), intervals.begin() + ib);
result.push_back(newInterval);
result.insert(result.end(), intervals.begin() + ie, intervals.end());
return result;
}
private:
static bool compare(const Interval& a, const Interval& b) { return a.start < b.start; }
};
void Print(const vector<Interval>& v) {
for (int i = 0; i < v.size(); ++i) {
cout << "[" << v[i].start << ", " << v[i].end << "] ";
}
cout << endl;
}
int main() {
Solution s;
// Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].
{
int b[] = {1, 6};
int e[] = {3, 9};
vector<Interval> intervals;
for (int i = 0; i < ARRAYSIZE(b); ++i) {
intervals.push_back(Interval(b[i], e[i]));
}
Interval i(2, 5);
Print(intervals);
Print(s.insert(intervals, i));
}
{
int b[] = {1};
int e[] = {5};
vector<Interval> intervals;
for (int i = 0; i < ARRAYSIZE(b); ++i) {
intervals.push_back(Interval(b[i], e[i]));
}
Interval i(2, 3);
Print(intervals);
Print(s.insert(intervals, i));
}
// Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].
{
int b[] = {1, 3, 6, 8, 12};
int e[] = {2, 5, 7, 10, 16};
vector<Interval> intervals;
for (int i = 0; i < ARRAYSIZE(b); ++i) {
intervals.push_back(Interval(b[i], e[i]));
}
Interval i(4, 9);
Print(intervals);
Print(s.insert(intervals, i));
}
}