182. Summary Ranges

Because the array is sorted, consecutive numbers form a run. Walk once: extend the current run while each next number is exactly one more than the last, then emit the run as either 'start' or 'start->end'.

Idea

The array is sorted and has no duplicates, so a maximal range is just a run of consecutive integers. We scan once, and whenever nums[i+1] == nums[i] + 1 we're still inside the same run; otherwise the run ends and we record it.

Algorithm

1. Start at index i = 0.
2. Mark start = nums[i]. Advance i while nums[i+1] == nums[i] + 1 (extend the run).
3. Now nums[i] is the run's end. Emit str(start) if start == nums[i], else str(start) + "->" + str(nums[i]).
4. Move past this run (i += 1) and repeat until the array is consumed.

Worked example: [0,1,2,4,5,7]

• start = 0; 1 = 0+1 ✓, 2 = 1+1 ✓, then 4 ≠ 2+1 → run is 0..2 → "0->2".
• start = 4; 5 = 4+1 ✓, then 7 ≠ 5+1 → run is 4..5 → "4->5".
• start = 7; no next → single → "7".

Result: ["0->2", "4->5", "7"].

Complexity

• Time: O(n) — one pass.
• Space: O(1) extra (ignoring the output list).

import java.util.ArrayList;
import java.util.List;

class Solution {
    public String[] summaryRanges(int[] nums) {
        List<String> result = new ArrayList<>();
        int i = 0, n = nums.length;
        while (i < n) {
            int start = nums[i];
            while (i + 1 < n && nums[i + 1] == nums[i] + 1) {
                i++;
            }
            if (start == nums[i]) {
                result.add(String.valueOf(start));
            } else {
                result.add(start + "->" + nums[i]);
            }
            i++;
        }
        return result.toArray(new String[0]);
    }
}