LeetCode Third Maximum Number 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.

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找出一个数组中的第三大的数，如果不存在第三大的数，则返回最大的数。数组中可能会有重复数字。要求用O(n)的时间。 定义最大的数、第二大、第三大的数分别为first、second、third，遍历数组然后不断更新这三个数，更新的方法是从最大数错位赋值，很容易理解。求数组中第二大的数也是类似的道理。 看代码就很好理解了，注意数组中可能会有INT_MIN，所以需要用long数据类型。另外，本题要求的第三大是相异数字的排名，比如第三个样例，第三大的应该是1，而不是2，虽然有第二个2排名第三，但和第二名重复了，不能算，这一点在两个else if中需要保证严格小于第一名或第二名。 [cpp] class Solution { public: int thirdMax(vector<int>& nums) { long first = LONG_MIN, second = LONG_MIN, third = LONG_MIN; 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_MIN)return first; else return third; } }; [/cpp] 本代码提交AC，用时6MS。]]>