LeetCode: Non-overlapping Intervals

LeetCode: Non-overlapping Intervals

Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

Note:

  1. You may assume the interval’s end point is always bigger than its start point.
  2. Intervals like [1,2] and [2,3] have borders “touching” but they don’t overlap each other.

Example 1:

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Input: [ [1,2], [2,3], [3,4], [1,3] ]

Output: 1

Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.

Example 2:

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Input: [ [1,2], [1,2], [1,2] ]

Output: 2

Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.

Example 3:

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Input: [ [1,2], [2,3] ]

Output: 0

Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
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/**
* 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; }
* }
*/
class Solution {
public int eraseOverlapIntervals(Interval[] intervals) {
Arrays.sort(intervals, Comparator.comparingInt(o -> o.start));
int result = 0;
int cursor = 0;
for (int i = 1; i < intervals.length; i++) {
Interval crt = intervals[i];
Interval last = intervals[cursor];
if (last.end > crt.start) {
result++;
if (last.end > crt.end) cursor = i;
} else {
cursor = i;
}
}
return result;
}
}