435. Non-overlapping Intervals

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:

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.

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.

一开始看题的时候，感觉是backtracking题目。选择overlap的任意一个移除，看结果。但是看了tag之后发现是greedy题目。后来一想，确实是，因为如果两个interval overlap移除end靠后那个就是最优。

需要remove掉overlap的，那么如果a，b overlap，我们就可以移除其中一个。移除的那个我们希望移除那个end 位置靠后的，因为移除那个，可能会减少需要移除的interval

 public class Solution {  
   public int eraseOverlapIntervals(Interval[] intervals) {  
     if(intervals.length == 0) return 0;  
     int n = intervals.length;  
     Arrays.sort(intervals, (a, b)->{  
       if(a.start == b.start) return Integer.compare(a.end, b.end);  
       return Integer.compare(a.start,b.start);  
     });  
     int ans = 0;  
     Interval interval = intervals[0];  
     for(int i = 1; i < n; i++){  
       if(overlap(interval, intervals[i])){  
         ans++;  
         //choose min end  
         if(interval.end > intervals[i].end)  
           interval = intervals[i];  
       }else  
         interval = intervals[i];  
     }  
     return ans;  
   }  
   public boolean overlap(Interval a, Interval b){  
     return (b.start < a.end &&a.start <= b.start)||(a.start < b.end && b.start <= a.start);  
   }  
 }