[435. Non-overlapping Intervals](https://leetcode.com/problems/non-overlapping-intervals/)

### 題目描述

Given an array of intervals `intervals` where `intervals[i] = [starti, endi]`, return *the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.*

### 範例

**Example 1:**

Input: intervals = [[1,2],[2,3],[3,4],[1,3]]
Output: 1
Explanation: [1,3] can be removed and the rest of the intervals are non-overlapping.

**Example 2:**

Input: intervals = [[1,2],[1,2],[1,2]]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of the intervals non-overlapping.

**Example 3:**

Input: intervals = [[1,2],[2,3]]
Output: 0
Explanation: You don't need to remove any of the intervals since they're already non-overlapping.

**Constraints**:

* 1 <= `intervals.length` <= 10^5^
* `intervals[i].length` == 2
* -5 * 10^4^ <= `starti` < `endi` <= 5 * 10^4^

### 解答
#### Javascript

function eraseOverlapIntervals(intervals) {
  intervals.sort((a, b) => a[1] - b[1]); // 依結束時間排序
  let count = 0;
  let end = intervals[0][1];

  for (let i = 1; i < intervals.length; i++) {
    const [curStart, curEnd] = intervals[i];
    if (curStart < end) {
      count++;
    } else {
      end = curEnd;
    }
  }

  return count;
}

> [name=Marsgoat][time=Jul 19, 2023]

#### Python

class Solution:
    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
        n = len(intervals)
        intervals.sort(key=lambda x: x[1])
        ans, pre = 0, float('-inf')
        for start, end in intervals:
            if start < pre:
                ans += 1
            else:
                pre = end
        return ans

> [name=Yen-Chi Chen][time=Wed, Jul 19, 2023]

### Reference

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