Approach

# 先排序再找original可能的值
Sort first, then determine the possible values of the original.

Time Complexity

# n = length of nums
O(nlogn)

Space Complexity

O(1)

Code

class Solution {
public:
    int findFinalValue(vector<int>& nums, int original) {
        sort(nums.begin(), nums.end());

        for (int i = 0 ; i < nums.size() ; i++) {
            if (nums[i] == original) {
                original *= 2;
            }
            else if (nums[i] >= original) {
                break;
            }
        }

        return original;
    }
};