Approach

# Array
紀錄出現的數字次數並比對

Time Complexity

O(n)

Space Complexity

O(n)

Code

class Solution {
public:
    int repeatedNTimes(vector<int>& nums) {
        int ele[10010] = {0}, cnt = 0, maxn = -1, ans = -1;

        for (int i = 0 ; i < nums.size() ; i++) {
            if (ele[nums[i]] == 0) {
                cnt++;
            }
            ele[nums[i]]++;
            maxn = max(maxn, nums[i]);
        }

        for (int i = 0 ; i <= maxn ; i++) {
            if (ele[i] == cnt - 1) {
                ans = i;
            }
        }

        return ans;
    }
};