[Array]414. Third Maximum Number

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.
思路：找出一个数组中第三大的元素，这里只是说integer，没有规定范围，因此应该考虑到int型的最大值和最小值，INT_MIN，INT_MAX,
自己的想法：（129ms）比较简单，使用额外的辅助空间，将不重复的元素排序

int thirdMax(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        vector<int> ans;
        int i;
        for(i=nums.size()-1;i>=0;--i)
        {
            if(find(ans.begin(),ans.end(),nums[i])==ans.end())//一开始自己的想法是进行排序，然后相邻元素是否相等，
                ans.push_back(nums[i]);
        }
        if(ans.size()<3)
            return ans[0];
        return ans[2];
    }

或者这样：

优秀代码：（6ms）既然要找出第三大的元素值，则首先设定三个最小值，然后遍历所有的元素，找出前三大的元素值

int thirdMax(vector<int>& nums) {
       long long  a, b, c;
        a = b = c = LONG_MIN  ;
        for(auto num : nums){
            if(num <= c || num ==b || num == a)
                continue;
            else 
                c = num;
            if(c > b)
                swap(b, c);
            if(b > a)
                swap(a, b);
        }
        return c == LONG_MIN ? a : c;
    }

函数set下的代码：

int thirdMax(vector<int>& nums) {
        set<int> top;
        for (auto i : nums) {
            top.insert(i);
            if (top.size() > 3) top.erase(top.begin());
        }
        return top.size() == 3 ? *top.begin() : *top.rbegin();
    }

 set参考链接：http://www.cnblogs.com/BeyondAnyTime/archive/2012/08/13/2636375.html