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题目描述:
给定一个非空数组,返回此数组中第三大的数。如果不存在,则返回数组中最大的数。要求算法时间复杂度必须是O(n)。
示例 1:
输入: [3, 2, 1] 输出: 1 解释: 第三大的数是 1.
示例 2:
输入: [1, 2] 输出: 2 解释: 第三大的数不存在, 所以返回最大的数 2 .
示例 3:
输入: [2, 2, 3, 1] 输出: 1 解释: 注意,要求返回第三大的数,是指第三大且唯一出现的数。 存在两个值为2的数,它们都排第二。
解题思路:首先要排序,还要去重。。直接用set。同时维护set的大小为3,最后头是第三大的数,如果大小不足3,则返回尾部的数
C++解法:
class Solution { public: int thirdMax(vector<int>& nums) { set<int> s; for(int num : nums){ s.insert(num); if(s.size() > 3){ s.erase(s.begin()); } } return s.size() == 3 ?*s.begin():*s.rbegin(); } };
The text was updated successfully, but these errors were encountered:
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题目描述:
给定一个非空数组,返回此数组中第三大的数。如果不存在,则返回数组中最大的数。要求算法时间复杂度必须是O(n)。
示例 1:
示例 2:
示例 3:
解题思路:首先要排序,还要去重。。直接用set。同时维护set的大小为3,最后头是第三大的数,如果大小不足3,则返回尾部的数
C++解法:
The text was updated successfully, but these errors were encountered: