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题目

Given an integer array nums, return the third distinct maximum number in this array. If the third maximum does not exist, return the maximum number.

Example 1:

Input: nums = [3,2,1]
Output: 1
Explanation:
The first distinct maximum is 3.
The second distinct maximum is 2.
The third distinct maximum is 1.

Example 2:

Input: nums = [1,2]
Output: 2
Explanation:
The first distinct maximum is 2.
The second distinct maximum is 1.
The third distinct maximum does not exist, so the maximum (2) is returned instead.

Example 3:

Input: nums = [2,2,3,1]
Output: 1
Explanation:
The first distinct maximum is 3.
The second distinct maximum is 2 (both 2’s are counted together since they have the same value).
The third distinct maximum is 1.

Constraints:

1 <= nums.length <= 10^4
-2^ 31 <= nums[i] <= 2^31 - 1

Follow up: Can you find an O(n) solution?

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代码

class Solution {public:int thirdMax(vector<int>& nums) {set<int> s;for(int i = 0; i<nums.size(); i++)s.insert(nums[i]);int res;set<int>::iterator iter;if(s.size() < 3)for(iter = s.begin(); iter != s.end(); iter++) res = *iter;else{int j = s.size() - 3;for(iter = s.begin(); iter != s.end() && j >= 0; iter++, j--) res = *iter;}return res;}
};

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题目

Example 1:

Example 2:

Example 3:

Constraints:

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代码

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