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Insert Interval
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applewjg committed Dec 14, 2014
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/*
Author: King, [email protected]
Date: 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.
* public class Interval {
* int start;
* int end;
* Interval() { start = 0; end = 0; }
* Interval(int s, int e) { start = s; end = e; }
* }
*/
public class Solution {
public List<Interval> insert_1(List<Interval> intervals, Interval newInterval) {
List<Interval> res = new ArrayList<Interval>();
boolean inserted = false;
for (Interval it : intervals) {
if (inserted || it.end < newInterval.start) {
res.add(it);
} else if (it.start > newInterval.end) {
res.add(newInterval);
res.add(it);
inserted = true;
} else {
newInterval.start = Math.min(newInterval.start, it.start);
newInterval.end = Math.max(newInterval.end, it.end);
}
}
if (inserted == false) res.add(newInterval);
return res;
}
public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
List<Interval> res = new ArrayList<Interval>();
int n = intervals.size();
int left = 0, right = n - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (intervals.get(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 - left) / 2;
if (intervals.get(mid).end < newInterval.end) left = mid + 1;
else right = mid - 1;
}
int idxEnd = left;
if (idxStart >= 0 && newInterval.start <= intervals.get(idxStart).end) {
newInterval.start = intervals.get(idxStart--).start;
}
if (idxEnd < n && newInterval.end >= intervals.get(idxEnd).start) {
newInterval.end = intervals.get(idxEnd++).end;
}
for (int i = 0; i <= idxStart; ++i) {
res.add(intervals.get(i));
}
res.add(newInterval);
for (int i = idxEnd; i < n; ++i) {
res.add(intervals.get(i));
}
return res;
}
}

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