Non-overlapping Intervals问题及解法

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本文介绍了一种通过贪心策略解决区间去重问题的方法。给定一系列区间，目标是最小化需要移除的区间数量，以确保剩余区间互不重叠。文章通过示例详细解释了算法的具体实现过程。

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问题描述：

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.

示例：

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.

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.

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

Output: 0

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

问题分析：

这里我们采用贪心策略，按照start排序，尽可能使两个相邻的interval不重叠，若是重叠，选择end小的那个interval留下，大的则remove掉。

过程详见代码：

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
class Solution {
public:
    int eraseOverlapIntervals(vector<Interval>& intervals) {
        sort(intervals.begin(), intervals.end(), 
			[](Interval& a, Interval&b){
			if (a.start != b.start) return a.start < b.start;
			else return a.end < b.end;
		});
		int res = 0,end = INT_MIN;
		for (int i = 0; i < intervals.size(); i++)
		{
			if (intervals[i].start >= end)
			{
				end = intervals[i].end;
			}
			else
			{
				if (intervals[i].end <= end)
				{
					end = end = intervals[i].end;
				}
				res++;
			}
		}
		return res;
    }
};