LeetCode-1262. Greatest Sum Divisible by Three

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

题解：

之前只想到O(n^2)解法，看讨论区解法：

使用两个数字维护目前为止数组中组成的余数为1和2的数字最小组成，判断最终结果如果正好整除、余数为1（减去数组中最小组成余数1的数字大小）、余数为2（减去数组中最小组成余数2的数字大小）。

该算法可以扩展到整除任意数字的最大值。

class Solution {
public:
    int maxSumDivThree(vector& nums) {
        int n = nums.size();
        if (n == 0) {
            return 0;
        }
        long dp = 0, dp1 = INT_MAX, dp2 = INT_MAX, sum = 0;
        for (int i = 0; i < n; i++) {
            if (nums[i] % 3 == 1) {
                dp2 = min(dp2, dp1 + long(nums[i]));
                dp1 = min(dp1, long(nums[i]));
            }
            if (nums[i] % 3 == 2) {
                dp1 = min(dp1, dp2 + long(nums[i]));
                dp2 = min(dp2, long(nums[i]));
            }
            sum += nums[i];
        }
        if (sum % 3 == 0) {
            return sum;
        }
        if (sum % 3 == 1) {
            return sum - dp1;
        }
        return sum - dp2;
    }
};