#228. Summary Ranges

Difficulty Level : Easy

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.

Solution

critical case : nums = [-2147483648,-2147483647,2147483647]
the input vector data type is int, so it is quite easy to fall into a trap if we are going to make A(int) - B(int) because the range of int is 2^32 ~ 2^32-1. So, we have to cast int into long type to guarantee the operation. Also, if we use a temporary vector, then it is easier to deal with the case when there is only one range using method of the vector.

#include <string>
#include <vector>

using namespace std;

class Solution {
private:
    vector<string> ret;
public:
    string printRanges(vector<string>& v){
        string buf; 
        if(v.size() > 1)
        {
            buf = v.front() + "->" + v.back();
        }
        else
        {
            buf = v.front();
        }
        return buf;
    }

    vector<string> summaryRanges(vector<int>& nums) {
        if (nums.empty()) return ret; // Handle empty input

        vector<string> tmp_v;
        tmp_v.push_back(to_string(nums[0]));

        for(int i = 1; i < nums.size(); i++) {
            if(static_cast<long long>(nums[i]) - static_cast<long long>(nums[i-1]) != 1) {
                ret.push_back(printRanges(tmp_v));
                tmp_v.clear();
                tmp_v.push_back(to_string(nums[i]));
            } 
            else
            {
                tmp_v.push_back(to_string(nums[i]));
            }
        }
        ret.push_back(printRanges(tmp_v)); // Add the last range to ret
        return ret;
    }
};