Leetcode 56. Merge Intervals & 57. Insert Interval

56. Merge Intervals

Total Accepted: 65710 Total
Submissions: 259254 Difficulty: Hard

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]

.

思路：

先按区间start排序，需要建一个新的class implements Comparator来排序。

然后取俩interval,如果前者的end大于后者的start，则新建一个interval，然后把prev变成这个新建的interval，进行下次循环；否则直接输出prev到结果中。

/**
* 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> merge(List<Interval> intervals) {
if(intervals==null || intervals.size()<=1) return intervals;
Collections.sort(intervals, new IntervalComparator());

ArrayList<Interval> result = new ArrayList<Interval>();

Interval prev = intervals.get(0);
for (int i = 1; i < intervals.size(); i++) {
Interval curr = intervals.get(i);

if (prev.end >= curr.start) {
// merged case
Interval merged = new Interval(prev.start, Math.max(prev.end, curr.end));
prev = merged;
} else {
result.add(prev);
prev = curr;
}
}
result.add(prev);
return result;
}

class IntervalComparator implements Comparator<Interval>{
public int compare(Interval i1, Interval i2){
return i1.start - i2.start;
}
}
}

57. Insert Interval

Total Accepted: 57111 Total
Submissions: 239980 Difficulty: Hard

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]

.

先把new interval插入给定的intervals里面，然后直接用56题代码，搞定！

/**
* 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(List<Interval> intervals, Interval newInterval) {
intervals.add(newInterval);
Collections.sort(intervals, new IntervalComparator());

ArrayList<Interval> result = new ArrayList<Interval>();

Interval prev = intervals.get(0);
for (int i = 1; i < intervals.size(); i++) {
Interval curr = intervals.get(i);

if (prev.end >= curr.start) {
// merged case
Interval merged = new Interval(prev.start, Math.max(prev.end, curr.end));
prev = merged;
} else {
result.add(prev);
prev = curr;
}
}
result.add(prev);
return result;
}

public class IntervalComparator implements Comparator<Interval>{
public int compare(Interval i1, Interval i2){
return i1.start - i2.start;
}
}
}