414.Third Maximum Number
414.Third Maximum Number
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: 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.
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
Last modified 11mo ago