update: 模板题4 - Conflicting Appointments

Author: LLancelot

merge-intervals.md

@@ -205,4 +205,36 @@ class Solution(object):
 
 - **Time complexity**: As we are iterating through both the lists once, the time complexity of the above algorithm is **O(N + M)**, where ‘N’ and ‘M’ are the total number of intervals in the input arrays respectively.
 
-- **Space complexity**: Ignoring the space needed for the result list, the algorithm runs in constant space **O(1)**.
+- **Space complexity**: Ignoring the space needed for the result list, the algorithm runs in constant space **O(1)**.
+
+
+
+## 模板题4 - Conflicting Appointments
+
+#### 题目
+
+Given an array of intervals representing ‘N’ appointments, find out if a person can **attend all the appointments**.
+
+**Example 1:**
+
+```
+Appointments: [[1,4], [2,5], [7,9]]
+Output: false
+Explanation: Since [1,4] and [2,5] overlap, a person cannot attend both of these appointments.
+```
+
+**Example 2:**
+
+```
+Appointments: [[6,7], [2,4], [8,12]]
+Output: true
+Explanation: None of the appointments overlap, therefore a person can attend all of them.
+```
+
+**Example 3:**
+
+```
+Appointments: [[4,5], [2,3], [3,6]]
+Output: false
+Explanation: Since [4,5] and [3,6] overlap, a person cannot attend both of these appointments.
+```