难度:Easy
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: 1Explanation: The third maximum is 1.Example 2:Input: [1, 2]Output: 2Explanation: The third maximum does not exist, so the maximum (2) is returned instead.Example 3:Input: [2, 2, 3, 1]Output: 1Explanation: Note that the third maximum here means the third maximum distinct number.Both numbers with value 2 are both considered as second maximum.
topK用最小堆,去重用set。
class Solution {public:int thirdMax(vector<int>& nums) {if(nums.size()<3) return *max_element(nums.begin(),nums.end());priority_queue<int,vector<int>,greater<int>>three;unordered_set<int>ele;for(int i=0;i<nums.size();i++){if(ele.count(nums[i])) continue;three.push(nums[i]);ele.insert(nums[i]);}if(three.size()==1) return three.top();if(three.size()==2) {three.pop(); return three.top();}while(three.size() !=3) three.pop();return three.top();}};
执行用时 :24 ms, 在所有 C++ 提交中击败了13.20%的用户 内存消耗 :10.8 MB, 在所有 C++ 提交中击败了5.05%的用户