Third Maximum Number问题及解法
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问题描述:
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).
示例:
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.
问题分析:
求解第三大的值,可利用C++中集合set的性质:实现了红黑树的平衡二叉检索树的数据结构,插入元素时,它会自动调整二叉树的排列,把元素放到适当的位置,以保证每个子树根节点键值大于左子树所有节点的键值,小于右子树所有节点的键值;另外,还得保证根节点左子树的高度与右子树高度相等。
若是把集合看成一个数组的话,那么他始终是按照从小到大的顺序排列。
过程详见代码:
class Solution {public: 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(); }};
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