## Question

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"
```

**Constraints:**

`0 <= nums.length <= 20`

`-231 <= nums[i] <= 231 - 1`

- All the values of
`nums`

are**unique**. `nums`

is sorted in ascending order.

## Algorithm

- Easy question, implement it using two pointers.
- A few situations to handle :
- Single number
- Followed with a range
- Followed with another single number

- Consecutive digits

- Single number
- Don't forget to handle the remaining string in the string builder when loop is done

## Code

class Solution { public List<String> summaryRanges(int[] nums) { List<String> res = new ArrayList<>(); if (nums == null || nums.length == 0) { return res; } int left = 0, right = 0; StringBuilder sb = new StringBuilder(); while (right < nums.length) { if (left == right) { sb = new StringBuilder(); sb.append(nums[right++]); } else if (nums[right] == nums[right - 1] + 1) { right++; } else { if (left != right - 1) { sb.append("->").append(nums[right - 1]); } res.add(sb.toString()); left = right; } } if (left != right - 1) { sb.append("->").append(nums[right - 1]); } res.add(sb.toString()); return res; } }