414. Third Maximum Number

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本文介绍了一种算法，用于从整数数组中找出第三大的不同数值。若数组中不足三个不同的数值，则返回最大值。该算法考虑了特殊情况并确保了O(n)的时间复杂度。

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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.

解题思路：若是数组中只出现一个或两个不同的元素，输出最大值。所以，设置count进行筛选。注意[1,-2147483648,2]和[1,2]这两种情况下，max都是2，second都是1，third都是-2147483648。但是前者应该输出-2147483648，后者应该输出2。所以需要t标志位来确定数组中是否出现-2147483648。

代码：

class Solution {
public:
int thirdMax(vector<int>& nums) {
int max, mid, min, count, t;
max = mid = min = Integer.MIN_VALUE;
count = 0;
t = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] == Integer.MIN_VALUE) {
t = 1;
}
if (nums[i] > max) {
min = mid;
mid = max;
max = nums[i];
count++;
}

else if (nums[i] > mid && nums[i] < max) {
min = mid;
mid = nums[i];
count++;
}

else if (nums[i] > min && nums[i] < mid) {
min = nums[i];
count++;
}
}
if (count < 2 || (count == 2 && t == 0)) {
return max;
} else {
return min;
}
}
};

不知道正确与否，望大家指正