leetcode 435. Non-overlapping Intervals

最新推荐文章于 2023-03-22 16:40:36 发布

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最新推荐文章于 2023-03-22 16:40:36 发布

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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移除区间数量的问题，以确保剩下的区间互不重叠。通过将问题类比为打气球，并采用排序和遍历策略，实现了高效求解。

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.

思路：

类似于打气球，先按照头排序，然后一个一个的遍历，如果不想交就更新区间，如果相交，end取较小的一个就可以。

bool cmp(const Interval& a,const Interval& b){
    return a.start<b.start?1:0;
}
class Solution {
public:
    int eraseOverlapIntervals(vector<Interval>& intervals) {
        int n = intervals.size();
        if(n == 0) return 0;
        sort(intervals.begin(),intervals.end(),cmp);
        int re = 0;
        int start = intervals[0].start, end = intervals[0].end, i = 1;    
        while(i<n){
            if(intervals[i].start>=end){
                end = intervals[i].end;
            } else {
                re++;
                if(intervals[i].end<end) end = intervals[i].end;
            }
            i++;
        }
        return re;
    }
};