You are given a sorted unique integer array nums.

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"

Example 3:

Input: nums = []
Output: []

Example 4:

Input: nums = [-1]
Output: ["-1"]

Example 5:

Input: nums = [0]
Output: ["0"]

Constraints:

0 <= nums.length <= 20
-2^31 <= nums[i] <= 2^31 - 1
All the values of nums are unique.

Solution

Time complexity : O(n)
Space complexity : O(1)

class Solution {
public:
    vector<string> summaryRanges(vector<int>& nums) {
        vector<string> ans = {};
        const int n = nums.size();
        
        for (int start=0, end=1; end<=n; ++end) {
            if (end == n || nums[end-1] != nums[end]-1) {
                if (start == end-1)
                    ans.push_back(to_string(nums[start]));
                else
                    ans.push_back(to_string(nums[start])+"->"+to_string(nums[end-1]));
                start = end;
            }
        }
        return ans;
    }
};

考慮目前的end與上一個是否差1，
若不是的話要切斷，
切斷時若start與end是同一個，就只要start即可，
切斷時若start與end是不同的，就要start->end-1。