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"
ifa != b
"a"
ifa == 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; } }