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:

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.

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.

這道題給了我們一堆區間，讓我們求需要至少移除多少個區間才能使剩下的區間沒有重疊，那麼我們首先要給區間排序，根據每個區間的start來做升序排序，然後我們開始要查詢重疊區間，判斷方法是看如果前一個區間的end大於後一個區間的start，那麼一定是重複區間，此時我們結果res自增1，我們需要刪除一個，那麼此時我們究竟該刪哪一個呢，為了保證我們總體去掉的區間數最小，我們去掉那個end值較大的區間，而在程式碼中，我們並沒有真正的刪掉某一個區間，而是用一個變數last指向上一個需要比較的區間，我們將last指向end值較小的那個區間；如果兩個區間沒有重疊，那麼此時last指向當前區間，繼續進行下一次遍歷，參見程式碼如下：

解法一：

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

我們也可以對上面程式碼進行簡化，主要利用三元操作符來代替if從句，參見程式碼如下： 

解法二：

class Solution {
public:
    int eraseOverlapIntervals(vector<Interval>& intervals) {
        if (intervals.empty()) return 0;      
        sort(intervals.begin(), intervals.end(), [](Interval& a, Interval& b){return a.start < b.start;});
        int res = 0, n = intervals.size(), endLast = intervals[0].end;
        for (int i = 1; i < n; ++i) {
            int t = endLast > intervals[i].start ? 1 : 0;
            endLast = t == 1 ? min(endLast, intervals[i].end) : intervals[i].end;
            res += t;
        }
        return res;
    }
};

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