228. Summary Ranges | zhengxingxing

Method 1

You are given a sorted unique integer array nums.
A range [a,b] is the set of all integers from a to b (inclusive).
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" if a != b
- "a" if a == 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"

Method 1

【O(n) time | O(1) space】

package Leetcode.GroupedLoop;

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

/**
 * @author zhengxingxing
 * @date 2025/04/13
 */
public class SummaryRanges {
    
    public static List<String> summaryRanges(int[] nums) {
        // Initialize result list to store range strings
        List<String> result = new ArrayList<>();

        // Handle edge cases: null array or empty array
        if (nums == null || nums.length == 0){
            return result;
        }

        // Get array length for iteration
        int n = nums.length;
        // Initialize index for traversal
        int i = 0;

        // Main loop to process the entire array
        while (i < n){
            // Mark the start of current group/range
            int start = i;

            // Inner loop to find the end of current continuous sequence
            // Continue while next number is consecutive
            while (i + 1 < n && nums[i + 1] == nums[i] + 1){
                i++;
            }

            // Process the found range
            if (start == i){
                // Case 1: Single number range (start equals end)
                // Format: "number"
                result.add(String.valueOf(nums[start]));
            }else{
                // Case 2: Multiple number range (start to end)
                // Format: "start->end"
                result.add(nums[start] + "->" + nums[i]);
            }
            // Move to the start of next potential range
            i++;
        }

        return result;
    }
    
    public static void main(String[] args) {
        // Test Case 1: Regular case with multiple ranges
        int[] nums1 = {0, 1, 2, 4, 5, 7};
        System.out.println("Test Case 1 Result: " + summaryRanges(nums1));
        // Expected output: ["0->2","4->5","7"]

        // Test Case 2: Mixed single numbers and ranges
        int[] nums2 = {0, 2, 3, 4, 6, 8, 9};
        System.out.println("Test Case 2 Result: " + summaryRanges(nums2));
        // Expected output: ["0","2->4","6","8->9"]

        // Test Case 3: Empty array edge case
        int[] nums3 = {};
        System.out.println("Test Case 3 Result: " + summaryRanges(nums3));
        // Expected output: []

        // Test Case 4: Single element array
        int[] nums4 = {1};
        System.out.println("Test Case 4 Result: " + summaryRanges(nums4));
        // Expected output: ["1"]
    }
}

Method 1

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