题目：
You are given a sorted unique integer array nums.

A range [a,b] is the set of all integers from a to b (inclusive).

Return the smallest sorted list of ranges that cover all the numbers in the array exactly. That is, each element of nums is covered by exactly one of the ranges, and there is no integer x such that x is in one of the ranges but not in nums.

Each range [a,b] in the list should be output as:

“a->b” if a != b
“a” if a == b

Example 1:

Input: nums = [0,1,2,4,5,7]
Output: [“0->2”,“4->5”,“7”]
Explanation: The ranges are:
[0,2] --> “0->2”
[4,5] --> “4->5”
[7,7] --> “7”
Example 2:

Input: nums = [0,2,3,4,6,8,9]
Output: [“0”,“2->4”,“6”,“8->9”]
Explanation: The ranges are:
[0,0] --> “0”
[2,4] --> “2->4”
[6,6] --> “6”
[8,9] --> “8->9”

Constraints:

0 <= nums.length <= 20
-231 <= nums[i] <= 231 - 1
All the values of nums are unique.
nums is sorted in ascending order.

解题思路：
我们从数组第一个数字开始，初始化一个变量start表示当前范围的起始值。
遍历数组中的每一个元素，如果当前元素 nums[i] 与前一个元素 nums[i-1] 不连续，那么我们将范围 [start, nums[i-1]] 添加到结果中，并更新 start 为当前元素 nums[i]。
遍历结束后，将最后一个范围添加到结果中。

python">class Solution:def summaryRanges(self, nums: List[int]) -> List[str]:result = []if not nums:return resultstart = nums[0]for i in range(1, len(nums)):if nums[i] != nums[i-1] + 1:if start == nums[i-1]:result.append(f"{start}")else:result.append(f"{start}->{nums[i-1]}")start = nums[i]if start == nums[-1]:result.append(f"{nums[-1]}")else:result.append(f"{start}->{nums[-1]}")return result

time complexity为O(n)。
注意，使用python3。