【LeetCode第 163 场周赛】5265. Greatest Sum Divisible by Three
【LeetCode第 163 场周赛】5265. Greatest Sum Divisible by Three
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5265. 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
来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/greatest-sum-divisible-by-three 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
动态规划，貌似我只会做这一道题，，，， ，
反正这个数被三除只有三种情况 整除 余1 余2，so只记录这三种情况最大和一直更新就OK啦
class Solution {
public:
int maxSumDivThree(vector<int>& nums) {
int one=0;
int two=0;
int three=0;
for(int i=0;i<nums.size();i++){
int a=nums[i]+one;
int b=nums[i]+two;
int c=nums[i]+three;
if(a%3==0){
if(a>three){
three=a;
}
}else if(a%3==1){
if(a>one){
one=a;
}
}else if(a%3==2){
if(a>two){
two=a;
}
}
if(b%3==0){
if(b>three){
three=b;
}
}else if(b%3==1){
if(b>one){
one=b;
}
}else if(b%3==2){
if(b>two){
two=b;
}
}
if(c%3==0){
if(c>three){
three=c;
}
}else if(c%3==1){
if(c>one){
one=c;
}
}else if(c%3==2){
if(c>two){
two=c;
}
}
//cout<<one<<" "<<two<<" "<<three<<endl;
}
return three;
}
};