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
class Solution {
    public int maxSumDivThree(int[] nums) {
        int res = 0, leftOne = 20000, leftTwo = 20000;
        for(int n:nums){
            res+=n;
            if(n%3==1){
                leftTwo = Math.min(leftTwo,leftOne+n);
                leftOne = Math.min(leftOne,n);
            }
            if(n%3==2) {
                leftOne = Math.min(leftOne,leftTwo+n);
                leftTwo = Math.min(leftTwo,n);
            }
        }
        if(res%3==0) return res;
        if(res%3==1) return res-leftOne;
        return res - leftTwo;
        
    }
}

https://leetcode.com/problems/greatest-sum-divisible-by-three/discuss/431108/Java-O(N)-solution-Simple-Math-O(1)-space

牛逼的方法,持续追踪除3余1和余2最小的数,最后总和整除就返回,余1就减去余1最小的数,余2同理。

注意无论余1还是余2都要持续更新另外一个,因为可能出现[2, 6, 2, 2, 7]这种情况,此时如果只追踪1或2,答案会是12而非15,因为2+2%3是1,7%3也是1,减的是7.

所以每次要更新两者。

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