leetcode435. Non-overlapping Intervals

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最新推荐文章于 2025-01-23 19:15:00 发布

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分类专栏： leetcode 文章标签： leetcode Greedy

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给定一组区间，找出需要删除的最少区间数，使得剩下的区间不重叠。按照区间end升序排列，如果下一个区间的start小于前一个的end，则删除下一个。解法包括两种策略。

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435. Non-overlapping Intervals

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.

解法一

按照end进行升序排列，如果下一个start小于上一个的end，则去除下一个。否则end变为下一个的end。

/**
 * 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;
        }

        Arrays.sort(intervals, new Comparator<Interval>() {
            @Override
            public int compare(Interval o1, Interval o2) {
                return o1.start - o2.start;
            }
        });
        int ret = 0;
        int terminal = intervals[0].end;
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i].start < terminal) {
                ret++;
                if (intervals[i].end < terminal) {
                    terminal = intervals[i].end;
                }
            } else {
                terminal = intervals[i].end;
            }
        }

        return ret;
    }
}

这里写图片描述

解法二

按照start升序排列，如果下一个的start比上一个的end小，就需要去除掉，如果此end比上一个end小，terminal变为小的end。如果下一个的start比上一个的end大，terminal扩展为现在的end。

/**
 * 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;
        }

        Arrays.sort(intervals, new Comparator<Interval>() {
            @Override
            public int compare(Interval o1, Interval o2) {
                return o1.start - o2.start;
            }
        });
        int ret = 0;
        int terminal = intervals[0].end;
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i].start < terminal) {
                ret++;
                if (intervals[i].end < terminal) {
                    terminal = intervals[i].end;
                }
            } else {
                terminal = intervals[i].end;
            }
        }

        return ret;

    }
}

这里写图片描述