Leetcode 1262. Greatest Sum Divisible by Three

Tyan

发布于 2021-08-06 15:27:45

3220

发布于 2021-08-06 15:27:45

文章被收录于专栏：SnailTyan

文章作者：Tyan 博客：noahsnail.com | CSDN | 简书

1. Description

2. Solution

**解析：**Version 1，是对Version 3的简化，具体参考Version 3。Version 2，先对数组求总和，然后根据每个数除3的余数进行有序分组，当总和除3余1时，此时总和应该减去最小的余数为1的数，或者减去最小的两个余数为2的数，如果二者同时有，则取二者的最小值，余数为2时同理。Version 3，dp用来维护最大和除3余数分别为0,1,2的状态，初始时，当余数为0时，不影响dp[0]的值，不需要更新dp[1],dp[2]，当余数为1,2时，此时才需要开始更新dp[1],dp[2]，然后根据余数的值不同，对各状态进行更新，pre用来保存上一轮结束时的状态。

Version 1

class Solution:
    def maxSumDivThree(self, nums: List[int]) -> int:
        n = len(nums)
        stat = [0, 0, 0]
        for i in range(n):
            for x in stat[:]:
                remainder = (x + nums[i]) % 3
                stat[remainder] = max(stat[remainder], x + nums[i])
        return stat[0]

Version 2

class Solution:
    def maxSumDivThree(self, nums: List[int]) -> int:
        total = sum(nums)
        n = len(nums)
        one = []
        two = []
        for i in range(n):
            if nums[i] % 3 == 1:
                bisect.insort(one, nums[i])
            elif nums[i] % 3 == 2:
                bisect.insort(two, nums[i])
        if total % 3 == 1:
            if len(one) == 0:
                total -= (two[0] + two[1])
            elif len(two) < 2:
                total -= one[0]
            else:
                total -= min(one[0], two[0] + two[1])
        elif total % 3 == 2:
            if len(two) == 0:
                total -= (one[0] + one[1])
            elif len(one) < 2:
                total -= two[0]
            else:
                total -= min(one[0] + one[1], two[0])
        return total

Version 3

class Solution:
    def maxSumDivThree(self, nums: List[int]) -> int:
        dp = [0, -float('inf'), -float('inf')]
        for num in nums:
            pre = dp[:]
            if num % 3 == 0:
                dp[0] = max(pre[0] + num, pre[0])
                dp[1] = max(pre[1] + num, pre[1])
                dp[2] = max(pre[2] + num, pre[2])
            elif num % 3 == 1:
                dp[0] = max(pre[2] + num, pre[0])
                dp[1] = max(pre[0] + num, pre[1])
                dp[2] = max(pre[1] + num, pre[2])
            else:
                dp[0] = max(pre[1] + num, pre[0])
                dp[1] = max(pre[2] + num, pre[1])
                dp[2] = max(pre[0] + num, pre[2])
        return dp[0]