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.
此题是meeting room的一个变型题目，要求最少数量的重叠部分，那么就是相当于用贪心选择算法求出最多可容纳的interval的数量，然后来检验是否有和它重叠的，如果有重叠的部分，就coun++，思路和meeting room是一模一样的。代码如下：
/**
* Definition for an interval.
* public class Interval {
* int start;
* int end;
* Interval() { start = 0; end = 0; }
* Interval(int s, int e) { start = s; end = e; }
* }
*/
public class Solution {
public int eraseOverlapIntervals(Interval[] intervals) {
if(intervals.length==0) return 0;
Arrays.sort(intervals,new Comparator<Interval>(){
public int compare(Interval a,Interval b){
if(a.end!=b.end){
return a.end-b.end;
}else{
return a.start-b.start;
}
}
});
int end = intervals[0].end;
int count = 0;
for(int i=1;i<intervals.length;i++){
if(intervals[i].start<end){
count++;
}else{
end = intervals[i].end;
}
}
return count;
}
}