LeetCode 414 Third Maximum Number

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。

 class Solution {
 public:
     static bool descending (int i, int j) { //自定义排序函数，此处应为static，否则报错
         return i > j;
     }

     int thirdMax(vector<int>& nums) {
         int len = nums.size();
         sort(nums.begin(), nums.end(), descending);
         //sort(nums.begin(). nums.end(), greater<int>());

         int j = ;
         for (int i = ; i < len; i++) {
             if(nums[i] < nums[i - ]) {
                 nums[j] = nums[i];
                 j++;
             }
         }

         return j >  ? nums[] : nums[];
     }
 };

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。

 class Solution {
 public:
     int thirdMax(vector<int>& nums) {
         int len = nums.size();
         set<int> s;
         for (int i = ; i < len; i++) {
             s.insert(nums[i]);
             if (s.size() > ) {
                 s.erase(s.begin());
             }
         }

         return s.size()== ? *s.begin() : *s.rbegin();
     }
 };

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.