leetcode解题报告（15）：Third Maximum Number

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).

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.

分析

强行用map写出来的，很粗暴的解法：

将nums的元素都放到map中，map的key是元素值，value是元素个数，放进去后map也被顺序排序好了。

如果排序后的元素个数大于等于3，那么第三大的元素就是map的最后一个迭代器减去3（map.end()指向的是尾后元素），如果小于3，返回最后一个元素即可。

代码如下：

class Solution {
public:
    int thirdMax(vector<int>& nums) {
        if(nums.size() == 0)return 0;
        map<int,int>m;
        for(int num : nums)
            ++m[num];
        if(m.size() == 1)   return m.begin()->first;
        else if(m.size() == 2)  return m.rbegin()->first;
        else{
            auto ret = m.rbegin();
            ++ret;
            ++ret;
            return ret->first;
        }
    }
};

在讨论区看到了一个用set解的，不过我的时间复杂度应该比这个更低: )

https://discuss.leetcode.com/topic/63903/short-easy-c-using-set

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();
    }
};