228. Summary Ranges

Description

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.

Submitted Code

class Solution(object):
    def summaryRanges(self, nums):
        """
        :type nums: List[int]
        :rtype: List[str]
        """
        if not nums:
            return nums

        result = []
        chain_start = nums[0]
        chain_end = nums[0]

        for n in nums[1:]:
            if n == chain_end + 1:
                chain_end = n
            else:
                if chain_start == chain_end:
                    result.append(str(chain_start))
                else:
                    result.append("{}->{}".format(chain_start, chain_end))
                chain_start = n
                chain_end = n

        if chain_start == chain_end:
            result.append(str(chain_start))
        else:
            result.append("{}->{}".format(chain_start, chain_end))
        
        return result

Runtime: 0 ms | Beats 100.00%
Memory: 12.46 MB | Beats 52.16%

두 개의 포인터를 이용하는 방법이다.

Other Solutions

1st

class Solution:
    def summaryRanges(self, nums: List[int]) -> List[str]:
        # Interval
        ret = []
        if not nums: return ret
        start = nums[0]
        for i in range(1, len(nums)):
            if nums[i] - nums[i-1] > 1:   # should split now
                line = str(start) if start == nums[i-1] else str(start) + "->" + str(nums[i-1])  # one num or more
                ret.append(line)
                start = nums[i]
        line = str(start) if start == nums[-1] else str(start) + "->" + str(nums[-1])  # add the last (n-1) case
        ret.append(line)
        return ret

time complexity: 𝑂(𝑛)
space complexity: 𝑂(𝑛)

하나의 포인터로도 해결할 수 있다.