統計刪除幾個區間使數組變爲無重疊的區間 Non-overlapping Intervals

問題：

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.

解決：

① 先排序，再刪除。根據每一個區間的start來作升序排序，而後咱們開始要查找重疊區間，判斷方法是看若是前一個區間的end大於後一個區間的start，那麼必定是重複區間，此時咱們結果res自增1，咱們須要刪除一個，那麼此時咱們究竟該刪哪個呢，爲了保證咱們整體去掉的區間數最小，咱們去掉那個end值較大的區間。

/**
 * 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; }
 * }
 */
class Solution { //5ms
    public int eraseOverlapIntervals(Interval[] intervals) {
        Arrays.sort(intervals, new Comparator<Interval>() {
            @Override
            public int compare(Interval o1, Interval o2) {//按照start升序排列
                return o1.start - o2.start;
            }
        });
        int count = 0;//記錄要刪除的個數
        int last = 0;
        int len = intervals.length;
        for (int i = 1;i < len;i ++){
            if (intervals[i].start < intervals[last].end){
                count ++;
                if (intervals[i].end < intervals[last].end) last = i;
            }else {
                last = i;
            }
        }
        return count;
    }
}