原题链接：Leetcode 414. Third Maximum Number
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?
方法一：遍历数组
思路：
题目要求返回第三大的数，以及当没有三个不同数的情况下返回第二个数
那么就维护三个值：first、second、third，最后分情况返回即可
比较麻烦的就是初始值，需要设置一个最小的正数，这里用 LONG_LONG_MIN
C++ 代码：
cpp
class Solution {
public:
// const long long Min = -(1LL << 31) - 1;
int thirdMax(vector<int>& nums) {
long first = LONG_LONG_MIN, second = LONG_LONG_MIN, third = LONG_LONG_MIN; // 当前最大的三个数
// 遍历的同时维护这三个数
for(int i = 0; i < nums.size(); i++ ){
if(nums[i] > first){
third = second;
second = first;
first = nums[i];
}
else if(nums[i] < first && nums[i] > second){
third = second;
second = nums[i];
}
else if(nums[i] < second && nums[i] > third){
third = nums[i];
}
}
if(third == LONG_LONG_MIN)
return first;
return third;
}
};