# [LeetCode] 435. Non-overlapping Intervals 不重叠的区间

## 题目

435. Non-overlapping Intervals

## 题意

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.


## 代码

/**
* Definition for an interval.
* type Interval struct {
*       Start int
*       End   int
* }
*/
func eraseOverlapIntervals(intervals []Interval) int {
if len(intervals) <= 0 {
return 0
}
sort.Slice(intervals, func(i, j int) bool {
return intervals[i].End < intervals[j].End
})
count := 1
end := intervals[0].End
for i:=1; i < len(intervals); i++ {
if intervals[i].Start < end {
continue
}
end = intervals[i].End
count++
}
return len(intervals) - count
}