LeetCode 228: Summary Ranges
LeetCode 228 Solution Explanation
Explanation
To solve this problem, we can iterate through the input array and keep track of the start and end of each range. Whenever there is a gap in the sequence, we add the current range to the result. We handle the cases where a range has a single number separately. Finally, we return the list of ranges as strings.
- Initialize an empty list to store the result ranges.
- Initialize variables
startandendto track the start and end of the current range. - Iterate through the input array.
- If the current number is consecutive with the previous number, update the
endto the current number. - If there is a gap, add the range [start, end] to the result list.
- Update
startto the current number. - Handle the case where a range has a single number.
- Add the final range to the result list.
- Convert the ranges to the required format and return.
Time Complexity: O(n) where n is the number of elements in the input array. Space Complexity: O(1) excluding the space required for the output.
LeetCode 228 Solutions in Java, C++, Python
class Solution {
public List<String> summaryRanges(int[] nums) {
List<String> result = new ArrayList<>();
if (nums == null || nums.length == 0) {
return result;
}
int start = nums[0];
int end = nums[0];
for (int i = 1; i < nums.length; i++) {
if (nums[i] == end + 1) {
end = nums[i];
} else {
if (start == end) {
result.add(Integer.toString(start));
} else {
result.add(start + "->" + end);
}
start = end = nums[i];
}
}
if (start == end) {
result.add(Integer.toString(start));
} else {
result.add(start + "->" + end);
}
return result;
}
}Interactive Code Editor for LeetCode 228
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