Leetcode 435. Non-overlapping Intervals

最新推荐文章于 2025-01-23 19:15:00 发布

原创最新推荐文章于 2025-01-23 19:15:00 发布 · 380 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

刷题专栏收录该内容

24 篇文章

订阅专栏

本文介绍了一种算法，用于解决给定一系列区间时如何最小化地移除某些区间以确保剩下的区间互不重叠的问题。通过排序和遍历，该算法能够高效地找到需要移除的区间数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

Example1:

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.

Example2:

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.

Example3:

Input: [ [1,2], [2,3] ]

Output: 0

Explanation: You don't need to remove any of the intervals since they're already non-overlapping.

思路

先对区间进行排序，然后遍历所有区间，尽量去掉长区间，则保留的区间数目一定最多。

代码

bool cmp(Interval a, Interval b) {
    return a.start < b.start; 
}

class Solution {
public:
    int eraseOverlapIntervals(vector<Interval>& intervals) {
        sort(intervals.begin(), intervals.end(), cmp);
        int res = 0, pre = 0; //res记录保留区间的个数，pre记录每次需要判断是否重叠的起始区间
        for (int i = 1; i < intervals.size(); i++) {
            if (intervals[i].start < intervals[pre].end) {
                res++; //当前区间起始值小于pre区间的末尾保留上一个区间
                if (intervals[i].end < intervals[pre].end) pre = i;
                //这个的意思是从集合中去掉的元素是pre（因为pre太长）
            }
            else pre = i; //区间i和区间pre已经不重叠，需要判断重叠的起始区间pre后移
        }
        return res;
    }
};