一、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. 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.

二、代码

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

三、代码解析

解题思路就是要先把所有区间按start的升序排序，然后去掉有重复的区间，然后运用贪心算法的思想，去掉end值较大的区间（因为end值越大，和后面的区间有重复的可能性越大）。这里的去掉不是真的删掉，而是改变last指向的位置，每一次用intervals[i]跟last指向的元素进行比较。