[Leetcode] 435. Non-overlapping Intervals

题目： leetcode链接
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:
* You may assume the interval’s end point is always bigger than its start point.
* 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.

注意的问题： 要删掉的区间应该是最少的。

思路： 将区间 按照end升序，当end相同时，按照start降序的顺序进行排序，这样排序的好处是 区间范围大的会排到后面，这样计算的是删掉最少的。

具体代码实现：

/**
 * 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 == 0){
            return 0;
        }
        int len = intervals.length;
        Arrays.sort(intervals, new Comparator() {

            public int compare(Interval o1, Interval o2) {
                if(o1.end != o2.end){
                    return o1.end - o2.end;
                }else{
                    return o2.start - o1.start;
                }
            }
        });

        int result = 0;
        int temp = intervals[0].end;
        int flag = 0;
        for(int i = 1; i < len; i++) {
            int num = intervals[i].start;
            if(num < temp){
                result++;
            }else{
                temp = intervals[i].end;
            }
        }

        return result;
    }
}

注：开始的想法是按照区间start升序排列 start相同时按照end升序排列，这样的排列方式虽然可以删掉重复区间，但是这种方式删掉的重复区间数，不是最少的，因为这样保留的区间不一定是覆盖范围小的，可能会把大区间留下
例子：[ [1,5], [2,3], [4,5] ]
这样的排列结果，在计算重复区间时，以第一个为基准，会把第二个和第三个区间删掉。这是错误的结果。