[LeetCode]414. Third Maximum Number第三大的数

知识点：最小整数，数组初始化vector，vector遍历

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).

给一个非空的整形数组，返回这个数组中第三大的数。如果它不存在，则返回最大的数。时间复杂度必须为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.

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void InsertOneNum2List(long sortList[3], int num)
    {	
        //sortList 0-max,1-second,2-third
        if(num == sortList[0] || num == sortList[1] ||num == sortList[2])
            return;
        if(num > sortList[0])
        {
            sortList[2] = sortList[1];
            sortList[1] = sortList[0];
            sortList[0] = num;
        }
        else if(num > sortList[1])
        {
            sortList[2] = sortList[1];
            sortList[1] = num;
        }
        else if(num > sortList[2])
        {
            sortList[2] = num;
        }
    }
    
class Solution {
public:
   

    int thirdMax(vector<int>& nums) {
        //因为会用INT_MIN作为测试数，所以这里需要用LONG_MIN
        long sortList[3] = {LONG_MIN,LONG_MIN,LONG_MIN};

//遍历插入
        vector<int>::iterator iter;
        for(iter = nums.begin();iter != nums.end();iter++)
        {
            InsertOneNum2List(sortList,*iter);
            
			
        }
    
        if(sortList[2] == LONG_MIN || sortList[2] == sortList[1])
            return sortList[0];
        else
            return sortList[2];
        }
};

void main()
{	
	int a[] = {1,2,-2147483648};
	vector<int> nums(a,a+3);
	Solution *s = new Solution();
	int result = s->thirdMax(nums);
	cout<<result;
}