# 435. Non-overlapping-Intervals

difficulty: Medium

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

1.

**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:**

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.

## Method One

```java
class Solution {
    public int eraseOverlapIntervals(int[][] intervals) {
        if(intervals.length < 2) {
            return 0;
        }
        Arrays.sort(intervals, (a,b) -> a[0] - b[0]);
        int count = -1;
        int curEnd = intervals[0][1];
        for(int[] interval : intervals) {
            if(interval[0] < curEnd) {
                count++;
                curEnd = Math.min( interval[1], curEnd );
            }else{
                curEnd = interval[1];
            }
        }
        return count;
    }
}
​
```
