163. Missing Ranges ----M

You are given an inclusive range [lower, upper] and a sorted unique integer array nums, where all elements are in the inclusive range.

A number x is considered missing if x is in the range [lower, upper] and x is not in nums.

Return the smallest sorted list of ranges that cover every missing number exactly. That is, no element of nums is in any of the ranges, and each missing number is in one of the ranges.

Each range [a,b] in the list should be output as:

• "a->b" if a != b
• "a" if a == b

 

Example 1:

Input: nums = [0,1,3,50,75], lower = 0, upper = 99
Output: ["2","4->49","51->74","76->99"]
Explanation: The ranges are:
[2,2] --> "2"
[4,49] --> "4->49"
[51,74] --> "51->74"
[76,99] --> "76->99"

Example 2:

Input: nums = [], lower = 1, upper = 1
Output: ["1"]
Explanation: The only missing range is [1,1], which becomes "1".

Example 3:

Input: nums = [], lower = -3, upper = -1
Output: ["-3->-1"]
Explanation: The only missing range is [-3,-1], which becomes "-3->-1".

Example 4:

Input: nums = [-1], lower = -1, upper = -1
Output: []
Explanation: There are no missing ranges since there are no missing numbers.

Example 5:

Input: nums = [-1], lower = -2, upper = -1
Output: ["-2"]

 

Constraints:

• -109 <= lower <= upper <= 109
• 0 <= nums.length <= 100
• lower <= nums[i] <= upper
• All the values of nums are unique.

A:
每看一个数字，表示之前的都处理完了。 trick： 把upper+1  放到最后

class Solution {
public:
    vector<string> findMissingRanges(vector<int>& nums, int lower, int upper) {
        int pre = lower-1;
        nums.push_back(upper+1);
        vector<string> res;
        for(auto end : nums){
            if(end != pre+1){
                res.push_back(getStr(pre+1, end-1));
            }
            pre = end;
        }
        nums.pop_back();// pop the newly added upper+1
        return res;
    }
private:
    string getStr(int a , int b){
        if(a==b){
            return to_string(a);
        }else{
            return to_string(a)+"->"+to_string(b);
        }
    }
};