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 * 10^41 <= 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; } }
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。