leetCode No.435 Non-overlapping Intervals

题目

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.

标签：Greedy

题意

给定一个Interval数组，每一个Interval对象包含一个start和end，代表一段长度，在整个数组中删除最少的节点，使得每个Interval对象代表的长度没有重合。

解题思路

按照每个节点的end从小到大排序，从第一个开始，将当前点的end和后面一个节点的start比较，如果end<=start那么说明没有重合，将后面一个节点作为当前点，继续向后比较。如果当前点的end和后面一个节点的start比较，end>start说明有重合，那么就跳过后面一个点，继续向后面比较。

代码

public class Solution {
    public int eraseOverlapIntervals(Interval[] intervals) {
        if (intervals == null || intervals.length == 0) {
            return 0;
        }
        Arrays.sort(intervals, new Comparator<Interval>() {

            @Override
            public int compare(Interval o1, Interval o2) {
                return o1.end - o2.end;
            }
        });
        int length = intervals.length;
        int count = 1;
        int last = 0;
        for (int i = 1; i < length; i++) {
            if (intervals[last].end <= intervals[i].start) {
                count ++;
                last = i;
            }
        }
        return length - count;

    }
}

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原题
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