# 435.Non-overlapping-Intervals

## 题目地址

https://leetcode.com/problems/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[1] - b[1];
});

int cnt = 1;
int end = intervals[0][1];
for (int i = 1; i < intervals.length; i++) {
if (intervals[i][0] >= end) {
cnt++;
end = intervals[i][1];
}
}

return intervals.length - cnt;
}
}``````

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