228. Summary Ranges

Given a sorted integer array without duplicates, return the summary of its ranges.

For example, given [0,1,2,4,5,7], return ["0->2","4->5","7"].

Solution 1, Extend the original array to include last number twice, this is deal-breaker.

public class Solution {
    public List<String> summaryRanges(int[] nums) {
        List<String> res = new ArrayList<>();
        if(nums == null || nums.length == 0) return res;
        int[] extend = new int[nums.length+1];
        for(int i=0; i< nums.length; i++){
            extend[i] = nums[i];
        }
        extend[nums.length] = extend[nums.length-1];
        int start = extend[0];
        for(int i=1; i< extend.length; i++){
            if(extend[i] != extend[i-1] +1){
                int end = extend[i-1];
                if(start == end){
                    res.add(Integer.toString(start));
                }else{
                    res.add(start + "->" + end);
                }

                start = extend[i];
            }
        }

        return res;

    }
}

without extending, special check in the ending index

public class Solution {
    public List<String> summaryRanges(int[] nums) {

        List<String> res = new ArrayList<>();
        if(nums == null || nums.length == 0) return res;

        int start = nums[0];
        for(int i=1; i<=nums.length; i++){
            if(i == nums.length){
                if(start == nums[nums.length-1]){
                    res.add(Integer.toString(start));
                }else{
                    res.add(start + "->" + nums[nums.length-1]);
                }
                break;
            }
            if(nums[i] != nums[i-1] +1){
                int end = nums[i-1];
                if(start == end){
                    res.add(Integer.toString(start));
                }else{
                    res.add(start + "->" + end);
                }
                start = nums[i];
            }
        }

        return res;
    }
}