LeetCode 228: Summary Ranges

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 start and end to 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 end to the current number.
• If there is a gap, add the range [start, end] to the result list.
• Update start to 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.

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;
    }
}