414 Third Maximum Number

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本文介绍了一种在非空整数数组中寻找第三大数的算法，并提供了具体实现案例。当数组中不存在第三大数时，则返回最大数。文章通过排序和去重实现了O(n)的时间复杂度。

题目来源【Leetcode】

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.

这道题比较简单，就是求第三大的数，考虑好几种特殊情况就没啥问题了

class Solution {
public:
    int thirdMax(vector<int>& nums) {
        int s = nums.size();
        if(s == 0) return 0;
        sort(nums.begin(),nums.end());
        nums.erase(unique(nums.begin(),nums.end()),nums.end());
        if(nums.size() == 1) return nums[0];
        else if(nums.size() == 2)  return nums[1];
        else if(nums.size() == 3)  return nums[0];
        else return nums[nums.size()-3];
    }
};