Non-overlapping Intervals

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本文介绍了一个算法，用于解决LeetCode上的Non-overlapping Intervals问题。该算法通过将区间按结束点排序并选择最早结束的区间，来找出最小数量的区间移除数，使剩余区间不重叠。文中提供了一个具体的C++实现示例。

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[leetcode]Non-overlapping Intervals

链接：https://leetcode.com/problems/non-overlapping-intervals/description/

Question

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.

Solution

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
// 思路和炸气球那个差不多，先对区间(x,y)按照y从小到大排序，然后确保第一个选择的情况下选择后面的，这样选出来的独立区间最大，最后减去这些独立的剩下的就是overlapping的
class Solution {
public:
  struct compare {
    bool operator()(const Interval& left, const Interval& right) {
      return left.end < right.end;
    }
  };
  int eraseOverlapIntervals(vector<Interval>& intervals) {
    if (intervals.size() == 0) return 0;
    sort(intervals.begin(), intervals.end(), compare());
    int myend = intervals[0].end;
    int cnt = 1;
    for (int i = 1; i < intervals.size(); i++) {
      if (intervals[i].start >= myend) {
        myend = intervals[i].end;
        cnt++;
      }
    }
    return intervals.size()-cnt;
  }
};

思路：具体看注释~