全组合会TLE，感觉是需要用DP，但是不会做。
找了一份参考答案，还在研究中。
1 class Solution: 2 def maxSumDivThree(self, nums: List[int]) -> int: 3 n = len(nums) 4 dp = [[0 for _ in range(n)] for _ in range(3)] 5 dp[nums[0] % 3][0] = nums[0] 6 for i in range(1, n): 7 for r in range(3): 8 include = dp[(r - nums[i]) % 3][i-1] + nums[i] 9 if include % 3 == r: 10 dp[r][i] = max(dp[r][i-1], include) 11 else: 12 dp[r][i] = dp[r][i-1] 13 return dp[0][n-1]
参考地址：https://leetcode.com/problems/greatest-sum-divisible-by-three/discuss/431105/Python-DP-solution
另外一个是使用数学方法解的：
1 class Solution(object): 2 def maxSumDivThree(self, nums): 3 dict1={} 4 sum1=sum(nums) 5 for i in nums: 6 g=i%3 7 if g in dict1: 8 dict1[g].append(i) 9 else: 10 dict1[g]=[i] 11 if 1 in dict1: 12 dict1[1]=sorted(dict1[1],reverse=True) 13 14 if 2 in dict1: 15 dict1[2]=sorted(dict1[2],reverse=True) 16 17 18 rem=sum1%3 19 if rem==0: 20 return sum1 21 22 if rem==1: 23 if rem in dict1: 24 max1=dict1[rem].pop() 25 max2=1000000000 26 if 2 in dict1 and len(dict1[2])>1: 27 max1=dict1[2].pop()+dict1[2].pop() 28 return sum1-min(max1,max2) 29 else: 30 sum1-=dict1[2].pop()+dict1[2].pop() 31 return sum1 32 else: 33 if rem in dict1: 34 max1=dict1[2].pop() 35 max2=1000000000 36 if 1 in dict1 and len(dict1[1])>1: 37 max2=dict1[1].pop()+dict1[1].pop() 38 return sum1-min(max1,max2) 39 else: 40 sum1-=dict1[1].pop()+dict1[1].pop() 41 return sum1
参考地址：https://leetcode.com/problems/greatest-sum-divisible-by-three/discuss/431078/Python-simple-solution
总结：这种DP和数学的题目，是大厂笔试时很喜欢出的题型，需要很强的数学思维(发现题目中隐藏的规律)。
我对这种题目束手无策，基本上一见到就放弃，只能看看大神们的解决方案，争取有所领悟吧。