leetcode Third Maximum Number 第三大的数

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).

题意：给定非空数组，找出第三大的数，若不存在，则输出最大数，要求时间复杂度位O(n)。

1、定义三个变量，分别存储前三大的数，初始化有漏洞，测试用例最小到INT_MIN，故初始化位LLONG_MIN。

class Solution {
public:
int thirdMax(vector<int>& nums) {
long long max = LLONG_MIN;
long long second = max;
long long third = max;
for (int i = 0; i < nums.size(); ++i) {
if (max < nums[i]) {
third = second;
second = max;
max = nums[i];
}
else if (second < nums[i] && max > nums[i]) {
third = second;
second = nums[i];
}
else if (third <= nums[i] && second > nums[i]) {
third = nums[i];
}
}
if (third != LLONG_MIN) return third;
return max;
}
};

2、利用set容器，自动排序

int thirdMax(vector<int>& nums) {
set<int> top3;
for (int num : nums) {
top3.insert(num);
if (top3.size() > 3)
top3.erase(top3.begin());
}
return top3.size() == 3 ? *top3.begin() : *top3.rbegin();
}