435. Non-overlapping Intervals

Problem:

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.

Analysis:

排序后从左往右扫，边扫边移除overlap。
一开始end = E1. 到了interval 2, 有overlap， 移除overlap, count++。不更新end, 因为interval 2 已经被移除了。

Solution:

class Solution {
    public int eraseOverlapIntervals(Interval[] intervals) {
        if ( intervals == null || intervals.length == 0)
            return 0;
        int res = 0, end = Integer.MIN_VALUE;
        Arrays.sort(intervals, ((a,b) -> a.end - b.end));
        for (int  i = 0; i < intervals.length; i++) {
            if (intervals[i].start >= end) {
                end = intervals[i].end;
            } else {
                res++;
            }
        }
        return res;
    }
}