Problem Description:

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.

NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

题解：

这个题是我算法课上的例题，原题是interval scheduling，

 

 

概括一下就是以结束时间排序，证明方法是反证法。时间复杂度为O(nlogn)，（排序）。

代码如下：

public int eraseOverlapIntervals(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> (a[1] - b[1]));
        Integer start = null;
        int res = 0;
        for(int i = 0; i < intervals.length; i++) {
            if(start == null || intervals[i][0] >= start) {
                start = intervals[i][1];
                res++;
            }
        }
        return intervals.length - res;
    }
}

 

 

 

 

 

 

 

LeetCode 435. Non-overlapping Intervals