Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

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.

Note:

  1. You may assume the interval’s end point is always bigger than its start point.
  2. Intervals like [1,2] and [2,3] have borders “touching” but they don’t overlap each other.

solution

Time complexity : O(nlogn)
Space complexity : O(1)

class Solution {
public:
    int eraseOverlapIntervals(vector<vector<int>>& intervals) {
        int n = intervals.size();

        sort(intervals.begin(), intervals.end(), 
             [](vector<int>& a, vector<int>& b) { return a[1] < b[1]; });

        int res = 0, x = 0;
        for (int i=1; i<n; ++i) {
            if (intervals[i][0] < intervals[x][1]) ++res;
            else x = i;
        }

        return res;
    }
};

首先對給定的時段，依據結束時間點小→大排序，會根據結束時間點排序的原因是，越早結束，越有多餘的時間排入下一個時間段。若用開始時間排序的話，有可能會得到一段很早開始、很晚結束的時間段。

再來是依據這個排序好的結果，取第一段做標記(x)，標記(x)的用意是代表那段是不會重疊的時間段。
從第二個時間段開始與x比較，其開始時間是否小於x？是的話即為重疊，重疊時間段數量結果+1，換下一位，不是的話即為未重覆且為剩餘時間段結束時間最早的，標記它為x，做到結束。