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This approach involves detecting the start and end of each range by iterating through the array sequentially. You remember the start of a potential range and adjust the range's end as long as consecutive numbers are found. When a break in consecutiveness occurs, you fix the end of the current range and start a new one.
Time Complexity: O(n)
Space Complexity: O(n)
1
2from typing import List
3
4def summaryRanges(nums: List[int]) -> List[str]:
5 ranges = []
6 i = 0
7 while i < len(nums):
8 start = i
9 while i + 1 < len(nums) and nums[i] + 1 == nums[i + 1]:
10 i += 1
11 if start == i:
12 ranges.append(str(nums[start]))
13 else:
14 ranges.append(f"{nums[start]}->{nums[i]}")
15 i += 1
16 return ranges
17Python uses a straightforward looping structure to find ranges. It uses format strings to easily concatenate numbers for ranges needing a '->'. This solution appends directly to list using range or single-potential numbers.
This approach utilizes a two-pointer method where one pointer marks the beginning of a new range, and another pointer (or the loop index itself) expands the range as far as possible until the next number isn't consecutive. Once a sequence ends, if numbers are the same, it is a single-element range; otherwise, a range connecting two different numbers is formed.
Time Complexity: O(n)
Space Complexity: O(n)
1#include <vector>
#include <string>
using namespace std;
vector<string> summaryRanges(vector<int>& nums) {
vector<string> result;
int n = nums.size();
int i = 0;
while (i < n) {
int start = i;
int j = i;
while (j + 1 < n && nums[j] + 1 == nums[j + 1]) {
j++;
}
if (start == j) {
result.push_back(to_string(nums[start]));
} else {
result.push_back(to_string(nums[start]) + "->" + to_string(nums[j]));
}
i = j + 1;
}
return result;
}
In C++, a nested loop designates two pointers to start and expand through the array, denoting where numbers are not consecutive anymore. It uses the starting and ending pointers to create a string of the range.