【leetcode】1262. Greatest Sum Divisible by Three

题目如下：

Given an array nums of integers, we need to find the maximum possible sum of elements of the array such that it is divisible by three.

Example 1:

Input: nums = [3,6,5,1,8]
Output: 18
Explanation: Pick numbers 3, 6, 1 and 8 their sum is 18 (maximum sum divisible by 3).

Example 2:

Input: nums = [4]
Output: 0
Explanation: Since 4 is not divisible by 3, do not pick any number.

Example 3:

Input: nums = [1,2,3,4,4]
Output: 12
Explanation: Pick numbers 1, 3, 4 and 4 their sum is 12 (maximum sum divisible by 3).

Constraints:

• 1 <= nums.length <= 4 * 10^4
• 1 <= nums[i] <= 10^4

解题思路：记sum为nums的和，如果其不能被3整除，那么余数只能是1或者2。如果余数是1，那么只需要减去nums中最小的除以3余数为1的数，或者减去nums中两个最小的除以3余数为2的数即可，两者之间选择较小的那个；如果余数是2，也同理。

代码如下：

class Solution(object):
    def maxSumDivThree(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        total = sum(nums)
        if total % 3 == 0:return total
        nums.sort()
        list_1 = []
        list_2 = []
        for i in nums:
            if i % 3 == 1 and len(list_1) < 2:
                list_1.append(i)
            elif i % 3 == 2 and len(list_2) < 2:
                list_2.append(i)
        if total % 3 == 1:
            minus = float('inf')
            if len(list_1) > 0:
                minus = list_1[0]
            if len(list_2) == 2:
                minus = min(minus,list_2[0] + list_2[1])
        elif total % 3 == 2:
            minus = float('inf')
            if len(list_2) > 0:
                minus = list_2[0]
            if len(list_1) == 2:
                minus = min(minus, list_1[0] + list_1[1])
        if minus == float('inf'):return 0
        return total - minus