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leetcode 414. 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: 1Explanation: The third maximum is 1.
Example 2:
Input: [1, 2]Output: 2Explanation: The third maximum does not exist, so the maximum (2) is returned instead.
Example 3:
Input: [2, 2, 3, 1]Output: 1Explanation: Note that the third maximum here means the third maximum distinct number.Both numbers with value 2 are both considered as second maximum.
求一堆数的前3大，一般排序就可以了，但是题目要求O(n)的时间去做
本题目也是6,
3 2 2 2 2 2 第三大是 3
2 2 2 第三大是2
没关系，这些题目中都已经说明了
class Solution {public: int thirdMax(vector<int>& nums) { int n = nums.size(); long Max = LONG_MIN; if (n < 3) { for (int i = 0; i < n; ++i) if (nums[i] > Max) Max = nums[i]; return Max; } else { long Max2 = LONG_MIN; long Max3 = LONG_MIN; for (int i = 0; i < n; ++i) { if (nums[i] > Max) Max3 = Max2 ,Max2 = Max, Max = nums[i]; else if (nums[i] > Max2 && nums[i] < Max) Max3 = Max2,Max2 = nums[i]; else if (nums[i] > Max3 && nums[i] < Max2) Max3 = nums[i]; } if (Max3 == LONG_MIN) Max3 = Max; return Max3; } }};
leetcode 414. Third Maximum Number
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