## [252\. Meeting Rooms](https://leetcode.com/problems/meeting-rooms/)

Given an array of meeting time `intervals` where `intervals[i] = [starti, endi]`, determine if a person could attend all meetings.

**Example 1:**

**Input:** intervals = \[\[0,30\],\[5,10\],\[15,20\]\]
**Output:** false

**Example 2:**

**Input:** intervals = \[\[7,10\],\[2,4\]\]
**Output:** true

**Constraints:**

- `0 <= intervals.length <= 104`
- `intervals[i].length == 2`
- `0 <= starti < endi <= 106`

算區間是否重疊，如果重疊 return false

class Solution {
public:
    bool canAttendMeetings(vector<vector<int>>& intervals) {
        int n = intervals.size();
        if (n == 0) return true;
        sort(intervals.begin(), intervals.end());
        for(int i = 0; i < n - 1; i++) {
            // 如果目前區間的 endtime 比下一個區間的 start time 大的話
            // 代表有 overlap
            if (intervals[i].back() > intervals[i + 1].front())
                return false;
        }
        return true;
    }
};

:::success
- 時間複雜度：$O(N \cdot \log N)$，取決於排序演算法
- 空間複雜度：$O(1)$
:::