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.

跟452. Minimum Number of Arrows to Burst Balloons类似，只需要改变一下计数的方式，这里出现overlap就计数++。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 == null || intervals.length <= 1) {            return 0;        }        Arrays.sort(intervals, new Comparator<Interval>(){            @Override            public int compare(Interval interval1, Interval interval2) {                return interval1.start != interval2.start? interval1.start - interval2.start: interval1.end - interval2.end;            }        });        int end = intervals[0].end, overlap = 0;        for (int i = 1; i < intervals.length; i ++) {            Interval temp = intervals[i];            if (temp.start < end) {                overlap ++;                end = Math.min(end, temp.end);            } else {                end = temp.end;            }        }        return overlap;    }}