228. Summary Ranges

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

Example 1:

Input:  [0,1,2,4,5,7]
Output: ["0->2","4->5","7"]
Explanation: 0,1,2 form a continuous range; 4,5 form a continuous range.

Example 2:

Input:  [0,2,3,4,6,8,9]
Output: ["0","2->4","6","8->9"]
Explanation: 2,3,4 form a continuous range; 8,9 form a continuous range.

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

双指针法，碉堡。貌似这种求首尾的题都可以用双指针法。

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

emmmm 我觉得这样表达更符合正常思维

总结：

we used two pointers to set a consecutive increament subarray. Everytime we get the start and end == i, then till there's not nums[i+1] == nums[i] + 1, we stop. Then if the end index didn't change, it means the next element is not consecutive. Otherwise, we add a summary of the subarray.