LeetCode solution summary 2178. Split into the sum of the largest number of positive and even numbers

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describe:

gives you an integer  finalSum . Please split it into the sum of several  positive even numbers that are different from each other  , and the number of positive even numbers split out  is the largest  .

• Let's say, for you  finalSum = 12 , then these splits are  satisfactory  (distinct positive even numbers and the sum is  finalSum): (2 + 10) , (2 + 4 + 6) and  (4 + 8) . Among them, (2 + 4 + 6) contains the largest number of integers. Note that  finalSum it cannot be split into  (2 + 2 + 4 + 4) , because the split integers must be different from each other.

Please return an integer array, which means splitting the integer into  the largest  number of positive even arrays. If there is no way to  finalSum split it, you return an  empty  array. You can return these integers in  any  order.

Example 1:

Input: finalSum = 12
 Output: [2,4,6]
 Explanation: Here are some splits that meet the requirements: sum (2 + 10)，(2 + 4 + 6) ( (4 + 8) 。
2 + 4 + 6) is the maximum number of integers, the number is 3, so we return [2,4 ,6]. 
[2,6,4] , [6,2,4] etc. are also feasible solutions.

Example 2:

Input: finalSum = 7
 Output: []
 Explanation: There is no way to split finalSum. 
So return empty array.

Example 3:

Input: finalSum = 28
 Output: [6,8,2,12]
 Explanation: Here are some splits that meet the requirements: The sum has the largest number of integers, which is 4, so we return [6,8,2,12 (2 + 26)，(6 + 8 + 2 + 12)] (4 + 24) 。
(6 + 8 + 2 + 12). 
[10,2,4,12] , [6,2,4,16] etc. are also feasible solutions.

hint:

• 1 <= finalSum <= 1010

Problem-solving ideas:

/**

* 2178. Split into the sum of the largest number of positive and even numbers

* Problem-solving ideas:

* First, judge whether it is an even number, if it is an odd number, return empty directly.

* Then, judge finalSum>=2*i+2, if the condition is met, it means that the remaining value is enough to put i and i+2, then finalSum -= i and list.push_back(i);

* Otherwise, the description is not enough, then directly add finalSum and jump out of the loop

*/

code:

class Solution2178
{
public:
    vector<long long> maximumEvenSplit(long long finalSum)
    {
        vector<long long> list;
        if (finalSum % 2 != 0)
        {
            return list;
        }
        for (long long i = 2;; i = i + 2)
        {
            // 保证下一轮还够
            if (finalSum - i >= i + 2)
            {
                finalSum -= i;
                list.push_back(i);
                continue;
            }
            // 如果下一轮不够，则直接使用剩余所有值
            list.push_back(finalSum);
            break;
        }
        return list;
    }
};