p435

LeetCode p435 Non-overlapping Intervals 题解

1.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:

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.

题意：

给你很多时间段，问最少剔除多少个，得到的段都不会有重叠。

2.解题思路:

贪心。
以终止点的大小从小到大排序，再累计所有没有重叠的个数。
具体见代码。

3.代码

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[title] [] [url] [link text]

 
/**
 * 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.length < 1)
			return 0;
		class myComparator implements Comparator<Interval> {
			public int compare(Interval a, Interval b) {
				return a.end - b.end; // 从小到大
			};
		}
		int count = 1;
		Arrays.sort(intervals, new myComparator()); //排序
		int end = intervals[0].end;
		for (Interval i : intervals) {
			if (i.start >= end) {
				end = i.end;
				count++;
			}
		}
		return intervals.length - count;
    }
}

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4.一些总结：