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: 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.

贪心法则，先把区间按照起点为第一关键字，终点为第二关键字排序。

排序以后挨个区间来考虑，当前为第K个区间，如果第K+1个区间和第K个区间有重叠，我们需要保留一个区间，

应该保留的区间是右端点值比较小的区间，因为区间已经按照起点排序过，保留右端点值较小的区间能够尽量地

减小和往后扫描到的区间有重叠的概率。

/** * 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<2)return 0;Arrays.sort(intervals,new Comparator<Interval>(){@Overridepublic int compare(Interval o1, Interval o2){// TODO Auto-generated method stubif(o1.start!=o2.start)return o1.start-o2.start;return o1.end-o2.end;}});Interval lastInterval=intervals[0];int cnt=0;for(int i=1;i<intervals.length;i++){if(intervals[i].start<lastInterval.end){cnt++;if(lastInterval.end<intervals[i].end)continue;}lastInterval=intervals[i];}return cnt;}}