Solution2.java

/// https://leetcode.com/problems/non-overlapping-intervals/description/
/// Author : liuyubobobo
/// Time   : 2017-11-19

import java.util.Arrays;
import java.util.Comparator;

/// Greedy Algorithm based on starting point
/// Time Complexity:  O(n)
/// Space Complexity: O(n)
public class Solution2 {

    // Definition for an interval.
    public static class Interval {
        int start;
        int end;
        Interval() { start = 0; end = 0; }
        Interval(int s, int e) { start = s; end = e; }
    }

    public int eraseOverlapIntervals(Interval[] intervals) {

        if(intervals.length == 0)
            return 0;

        Arrays.sort(intervals, new Comparator<Interval>() {
            @Override
            public int compare(Interval o1, Interval o2) {
                if(o1.start != o2.start)
                    return o1.start - o2.start;
                return o1.end - o2.end;
            }
        });

        int res = 1;
        int pre = 0;
        for(int i = 1 ; i < intervals.length ; i ++)
            if(intervals[i].start >= intervals[pre].end){
                res ++;
                pre = i;
            }
            else if(intervals[i].end < intervals[pre].end)
                pre = i;

        return intervals.length - res;
    }

    public static void main(String[] args) {
        Interval[] interval1 = {new Interval(1,2),
                new Interval(2,3),
                new Interval(3,4),
                new Interval(1,3)};
        System.out.println((new Solution2()).eraseOverlapIntervals(interval1));

        Interval[] interval2 = {new Interval(1,2),
                new Interval(1,2),
                new Interval(1,2)};
        System.out.println((new Solution2()).eraseOverlapIntervals(interval2));

        Interval[] interval3 = {new Interval(1,2),
                new Interval(2,3)};
        System.out.println((new Solution2()).eraseOverlapIntervals(interval3));
    }
}