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# 435.Non-overlapping-Intervals

## 题目描述

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

## 代码

### Approach #1

Time: O(nlogn) && Space: O(1)
class Solution {
public int eraseOverlapIntervals(int[][] intervals) {
if (intervals.length == 0) return 0;
Arrays.sort(intervals, (a, b) -> {
return a - b;
});
int cnt = 1;
int end = intervals;
for (int i = 1; i < intervals.length; i++) {
if (intervals[i] >= end) {
cnt++;
end = intervals[i];
}
}
return intervals.length - cnt;
}
}