[leetcode]435. Non-overlapping Intervals

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本文探讨了一种算法问题：如何移除最少数量的区间使剩余区间不再重叠。通过实例展示了不同输入情况下的解决方案，并提供了一段C++代码实现，利用贪心算法思想，以区间的结束点作为主要排序依据。

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.

讲真，一点都不想写这个的题解

被自己蠢哭了，告诉我这个题和看电视节目那个贪心入门有什么区别？？？

怎么还能第一关键字是start？？？后一个和前一个比较的是end，当然是第一关键字是end排序啊！！！

大二下开学还是老陈讲的啊………………

这都能写错，还搜题解…………

赶紧老实儿的一天刷一个题吧

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
bool comp(const Interval &a,const Interval &b)
    {
        if(a.end==b.end)return a.start<b.start;
        return a.end<b.end;
    }
class Solution {
public:
    
    int eraseOverlapIntervals(vector<Interval>& intervals) {
        if(intervals.size()==0) return 0;
        sort(intervals.begin(),intervals.end(),comp);
        int num=1;
        int right=intervals[0].end;
        for(int i=1;i<intervals.size();i++)
        {
            if(intervals[i].start<right) continue;
            num++;
            right=intervals[i].end;
        }
        return  intervals.size()-num;
    }
};