Date and Time: Sep 6, 2024, 15:10 (EST)
Link: https://leetcode.com/problems/summary-ranges/
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"
ifa != b
-
"a"
ifa == 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"
-
0 <= nums.length <= 20
-
-2^31 <= nums[i] <= 2^31 - 1
-
All the values of
nums
are unique. -
nums
is sorted in ascending order.
-
First we check if
nums
is empty or not. -
We use
start
to keep track of the first element of the continue interval, ifnums[i] != nums[i-1]+1
, we knownums[i]
is not part of the previous interval, so we can add this interval[start, nums[i-1]]
intores
. However, we have two cases to check, ifstart == nums[i-1]
, the previous value should be added tores
by itself. Otherwise, we add the intervalstr(start) + "->" + str(nums[i-1])
. After that, we updatestart = nums[i]
. -
We handle the last element in
nums
, we repeat to check ifstart == nums[-1]
, we add the last element by itself tores[]
, otherwise, we addstr(start) + "->" + str(nums[i-1])
.
class Solution:
def summaryRanges(self, nums: List[int]) -> List[str]:
if not nums:
return []
res = []
start = nums[0]
for i in range(1, len(nums)):
if nums[i] != nums[i-1]+1:
if start == nums[i-1]:
res.append(str(start))
else:
res.append(str(start) + "->" + str(nums[i-1]))
start = nums[i]
if start == nums[-1]:
res.append(str(start))
else:
res.append(str(start) + "->" + str(nums[-1]))
return res
Time Complexity: n
is length of nums
.
Space Complexity: