LeetCode : 163. Missing Ranges

[congr]

https://leetcode.com/problems/missing-ranges/

[congr]

O(N), very tricky - watch out overflow cases

public class Solution {
   public List<String> findMissingRanges(int[] nums, int lower, int upper) {
        List<String> res = new ArrayList<>();

        long prev = (long)lower - 1; // int -> long !!!
        long curr = 0;

        for(int i = 0; i <= nums.length; i++) {
            curr = i == nums.length ? (long)upper + 1: (long)nums[i]; // int -> long
            if(curr - prev == 2) res.add((prev + 1) + "");
            else if(curr - prev > 2) res.add((prev + 1) + "->" + (curr - 1));

            prev = curr;
        }

        return res;
    }
}

[congr]

[congr]

The first accepted code. 0ms

class Solution {
    public List<String> findMissingRanges(int[] nums, int lower, int upper) {
        List<String> list = new ArrayList();
        if (nums.length == 0) {
            String t = range((long)upper+1, (long)lower-1);
            if (!t.isEmpty()) list.add(t);
            return list;
        }

        String t = range((long)nums[0], (long)lower-1);
        if (!t.isEmpty()) list.add(t);

        for (int i = 1; i<nums.length; i++) {
            t = range((long)nums[i], (long)nums[i-1]); // a, b
            if (!t.isEmpty()) list.add(t); 
        }    

        t = range((long)upper+1, (long)nums[nums.length-1]);
        if (!t.isEmpty()) list.add(t);

        return list;
    }

    // [2147483647] 0 2147483647

    String range(long a, long b) { // a>b
        if (a - b == 2) 
            return String.valueOf(b + 1);
        else if (a - b > 2) 
            return String.valueOf((b + 1) + "->" + (a - 1));
        else 
            return "";
    }
}