Problem
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:
You may assume the interval's end point is always bigger than its start point.
Intervals like [1,2] and [2,3] have borders "touching" but they don't overlap each other.
Example 1:
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:
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:
Input: [ [1,2], [2,3] ]
Output: 0
Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
Solution
class Solution {
public int eraseOverlapIntervals(Interval[] intervals) {
if (intervals == null || intervals.length == 0) return 0;
//find non-overlapping count, so sort by end
Arrays.sort(intervals, (a, b)->(a.end-b.end));
int end = intervals[0].end;
int count = 1;
for (int i = 1; i < intervals.length; i++) {
if (intervals[i].start >= end) {
end = intervals[i].end;
count++;
}
}
return intervals.length-count;
}
}
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