LeetCode 414 Third Maximum Number

Posted 2020-08-25 SillyVicky

Problem:

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.

Summary:

找出整形数组中第三大的数，若存在第三大的数则返回该数，否则返回最大的数。

注意所谓的“大”为严格意义上的大于，大于等于的关系不成立。

Solution:

1. 首先将数组从大到小排序，然后从头遍历去掉重复数字，同时用一个指针记录当下去掉重复数字后的索引值。最后判断新数组中数字个数是否大于3即可。

 1 class Solution {
 2 public:
 3     static bool descending (int i, int j) { //自定义排序函数，此处应为static，否则报错
 4         return i > j;
 5     }
 6     
 7     int thirdMax(vector<int>& nums) {
 8         int len = nums.size();
 9         sort(nums.begin(), nums.end(), descending);
10         //sort(nums.begin(). nums.end(), greater<int>());
11         
12         int j = 1;
13         for (int i = 1; i < len; i++) {
14             if(nums[i] < nums[i - 1]) {
15                 nums[j] = nums[i];
16                 j++;
17             }
18         }
19         
20         return j > 2 ? nums[2] : nums[0];
21     }
22 };

2. 用set维护当前的三个最大值，由于set有去重和自动从小到大排序的功能，可以再size大于3时去掉最小的，即当前的第四大数。

 1 class Solution {
 2 public:
 3     int thirdMax(vector<int>& nums) {
 4         int len = nums.size();
 5         set<int> s;
 6         for (int i = 0; i < len; i++) {
 7             s.insert(nums[i]);
 8             if (s.size() > 3) {
 9                 s.erase(s.begin());
10             }
11         }
12         
13         return s.size()==3 ? *s.begin() : *s.rbegin();
14     }
15 };

参考：http://www.cnblogs.com/grandyang/p/5983113.html

 

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.