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[LeetCode Road] Summary Ranges - Solution/C++

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228. Summary Ranges

Question:

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:

    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"

    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"

    Input: nums = []
    Output: []

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

    Constraints:
  • 0 <= nums.length <= 20
  • -231 <= nums[i] <= 231 - 1
  • All the values of nums are unique.
  • nums is sorted in ascending order.

    • Source code

    Version 1

    Idea:
    This question is very easy, Use one variable i to record the head of valid range. i will be added 1 while the current value is not same as the next value. On the contrary, there are two sutiations to store the results. One is i = j and the other is i != j.

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

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