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.
思路
要找到最少需要移除多少个区间,找到不重叠区间最长可以是多少即可。
如果构成一个不重叠区间的最后一个区间的end
是当前最小的,那么,留给后续区间的空间越大。既而,能选择的区间数就更多。
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
}
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