Given a non-empty array of integers, return the third maximum number in this array. If it does not exist, return the maximum number. The time complexity must be in O(n).

Example 1:
Input: [3, 2, 1]

Output: 1

Explanation: The third maximum is 1.
Example 2:
Input: [1, 2]

Output: 2

Explanation: The third maximum does not exist, so the maximum (2) is returned instead.
Example 3:
Input: [2, 2, 3, 1]

Output: 1

Explanation: Note that the third maximum here means the third maximum distinct number.
Both numbers with value 2 are both considered as second maximum.

題目連結

找到第三大的數字 時間複雜度是O（n）
考慮遍歷三次：

但是會出現，其中出現最小數的情況，然後程式碼就跟別人不一樣了。。。。

我考慮標記位置來紀錄，因為位置不能是負數

class Solution {
public:
    int thirdMax(vector
 
<int>& nums) {
        int len=nums.size();

        int maxnum=nums[0];
        int maxpos=0;
        for(int i=0;i<len;i++){
            if(maxnum<nums[i]) {
                maxnum=nums[i];
                maxpos=i;
            }
        }

        int second=INT_MIN;
        int secpos=-1
 
;

        for(int i=0;i<len;i++){
            if(nums[i]==maxnum) continue;
            if(second<=nums[i]) {
                second=nums[i];
                secpos=i;
            }
        }

        if(second==INT_MIN && secpos==-1) {
            return maxnum;
        }

        int third=INT_MIN;
        int thpos=-1;

        for(int i=0;i<len;i++){
            if(nums[i]==maxnum || nums[i]==second) continue;
            if(third<=nums[i]) {
                third=nums[i];
                thpos=i;
            }
        }

        return (third==INT_MIN && thpos==-1) ? maxnum:third;
    }
};

程式碼有點長，學習別人的：

class Solution {
public:
    int thirdMax(vector<int>& nums) {
        long first = LONG_MIN, second = LONG_MIN, third = LONG_MIN;
        for (int num : nums) {
            if (num > first) {
                third = second;
                second = first;
                first = num;
            } else if (num > second && num < first) {
                third = second;
                second = num;
            } else if (num > third && num < second) {
                third = num;
            }
        }
        return (third == LONG_MIN || third == second) ? first : third;
    }
};