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

**Example 2:**

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

**Example 3:**

Input:[[1,2],[2,3]]Output:0Explanation: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.

```
struct Solution;
impl Solution {
fn erase_overlap_intervals(mut intervals: Vec<Vec<i32>>) -> i32 {
if intervals.is_empty() {
return 0;
}
let n = intervals.len();
intervals.sort_by_key(|v| v[1]);
let mut end = intervals[0][1];
let mut res = 0;
for i in 1..n {
if intervals[i][0] < end {
res += 1;
} else {
end = intervals[i][1];
}
}
res
}
}
#[test]
fn test() {
let intervals = vec_vec_i32![[1, 2], [2, 3], [3, 4], [1, 3]];
let res = 1;
assert_eq!(Solution::erase_overlap_intervals(intervals), res);
let intervals = vec_vec_i32![[1, 2], [1, 2], [1, 2]];
let res = 2;
assert_eq!(Solution::erase_overlap_intervals(intervals), res);
let intervals = vec_vec_i32![[1, 2], [2, 3]];
let res = 0;
assert_eq!(Solution::erase_overlap_intervals(intervals), res);
}
```