Summary Ranges You are given a sorted unique integer array nums
.
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"
Constraints:
0 <= nums.length <= 20
-231 <= nums[i] <= 231 - 1
- All the values of
nums
are unique. nums
is sorted in ascending order.
Summary Ranges Solutions
✅Time: O(n)
✅Space: O(n)
C++
class Solution {
public:
vector<string> summaryRanges(vector<int>& nums) {
vector<string> ans;
for (int i = 0; i < nums.size(); ++i) {
const int begin = nums[i];
while (i + 1 < nums.size() && nums[i] == nums[i + 1] - 1)
++i;
const int end = nums[i];
if (begin == end)
ans.push_back(to_string(begin));
else
ans.push_back(to_string(begin) + "->" + to_string(end));
}
return ans;
}
};
Java
class Solution {
public List<String> summaryRanges(int[] nums) {
List<String> ans = new ArrayList<>();
for (int i = 0; i < nums.length; ++i) {
final int begin = nums[i];
while (i + 1 < nums.length && nums[i] == nums[i + 1] - 1)
++i;
final int end = nums[i];
if (begin == end)
ans.add("" + begin);
else
ans.add("" + begin + "->" + end);
}
return ans;
}
}
Python
class Solution:
def summaryRanges(self, nums: List[int]) -> List[str]:
ans = []
i = 0
while i < len(nums):
begin = nums[i]
while i < len(nums) - 1 and nums[i] == nums[i + 1] - 1:
i += 1
end = nums[i]
if begin == end:
ans.append(str(begin))
else:
ans.append(str(begin) + "->" + str(end))
i += 1
return ans
Watch Tutorial
Checkout more Solutions here