Question

Formatted question description: https://leetcode.ca/all/228.html

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

Traverse the array once and check whether the next number is increasing each time

If yes, continue to traverse down
If not, we have to judge whether it is a number or a sequence at this time,
- A number is directly stored in the result
- The sequence should be stored in the first and last numbers and the arrow "->".

We need two variables i and j, where i is the position of the starting number of the continuous sequence, j is the length of the continuous sequence,

when j is 1, it means there is only one number,
if it is greater than 1, it is a continuous sequence.

Code

import java.util.ArrayList;
import java.util.List;

public class Summary_Ranges {

    class Solution {
        public List<String> summaryRanges(int[] nums) {
            List<String> result = new ArrayList<>();
            int left = 0, n = nums.length;
            while (left < n) {
                int right = 1;
                while (left + right < n && (long)nums[left + right] - nums[left] == right) {
                    ++right;
                }
                result.add(
                    right <= 1 ? // note: right here is already after ++ in above loop. so it's actually checking if it's a single digit
                        String.valueOf(nums[left])
                        : String.valueOf(nums[left]) + "->" + String.valueOf(nums[left + right - 1])); // also here, -1 because already done ++
                left += right;
            }

            return result;

        }
    }
}

############

class Solution {
    public List<String> summaryRanges(int[] nums) {
        List<String> ans = new ArrayList<>();
        for (int i = 0, j, n = nums.length; i < n; i = j + 1) {
            j = i;
            while (j + 1 < n && nums[j + 1] == nums[j] + 1) {
                ++j;
            }
            ans.add(f(nums, i, j));
        }
        return ans;
    }

    private String f(int[] nums, int i, int j) {
        return i == j ? nums[i] + "" : String.format("%d->%d", nums[i], nums[j]);
    }
}

// OJ: https://leetcode.com/problems/summary-ranges/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<string> summaryRanges(vector<int>& A) {
        vector<string> ans;
        int start = 0;
        for (int i = 0, N = A.size(); i < N; ++i) {
            if (i + 1 < N && A[i + 1] == A[i] + 1) continue;
            ans.push_back(to_string(A[start]) + (i == start ? "" : ("->" + to_string(A[i]))));
            start = i + 1;
        }
        return ans;
    }
};

class Solution:
    def summaryRanges(self, nums: List[int]) -> List[str]:
        def f(i, j):
            return str(nums[i]) if i == j else f'{nums[i]}->{nums[j]}'

        i = 0
        n = len(nums)
        ans = []
        while i < n:
            j = i
            while j + 1 < n and nums[j + 1] == nums[j] + 1:
                j += 1
            ans.append(f(i, j))
            i = j + 1
        return ans

############

class Solution(object):
  def summaryRanges(self, nums):
    """
    :type nums: List[int]
    :rtype: List[str]
    """

    def outputRange(start, end):
      if start == end:
        return str(start)
      return "{}->{}".format(start, end)

    if not nums:
      return []
    ans = []
    start = 0
    for i in range(0, len(nums) - 1):
      if nums[i] + 1 != nums[i + 1]:
        ans.append(outputRange(nums[start], nums[i]))
        start = i + 1
    ans.append(outputRange(nums[start], nums[-1]))
    return ans

func summaryRanges(nums []int) (ans []string) {
	f := func(i, j int) string {
		if i == j {
			return strconv.Itoa(nums[i])
		}
		return strconv.Itoa(nums[i]) + "->" + strconv.Itoa(nums[j])
	}
	for i, j, n := 0, 0, len(nums); i < n; i = j + 1 {
		j = i
		for j+1 < n && nums[j+1] == nums[j]+1 {
			j++
		}
		ans = append(ans, f(i, j))
	}
	return
}

public class Solution {
    public IList<string> SummaryRanges(int[] nums) {
        var ans = new List<string>();
        for (int i = 0, j = 0, n = nums.Length; i < n; i = j + 1) {
            j = i;
            while (j + 1 < n && nums[j + 1] == nums[j] + 1) {
                ++j;
            }
            ans.Add(f(nums, i, j));
        }
        return ans;
    }

    public string f(int[] nums, int i, int j) {
        return i == j ? nums[i].ToString() : string.Format("{0}->{1}", nums[i], nums[j]);
    }
}

function summaryRanges(nums: number[]): string[] {
    const f = (i: number, j: number): string => {
        return i === j ? `${nums[i]}` : `${nums[i]}->${nums[j]}`;
    };
    const n = nums.length;
    const ans: string[] = [];
    for (let i = 0, j = 0; i < n; i = j + 1) {
        j = i;
        while (j + 1 < n && nums[j + 1] === nums[j] + 1) {
            ++j;
        }
        ans.push(f(i, j));
    }
    return ans;
}

228 - Summary Ranges

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