[985] Sum of Even Numbers After Queries

We have an array A of integers, and an array queries of queries.

For the i-th query val = queries[i][0], index = queries[i][1], we add val to A[index].  Then, the answer to the i-th query is the sum of the even values of A.

(Here, the given index = queries[i][1] is a 0-based index, and each query permanently modifies the array A.)

Return the answer to all queries.  Your answer array should have answer[i] as the answer to the i-th query.

Example 1:

Input: A = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]
Output: [8,6,2,4]
Explanation: 
At the beginning, the array is [1,2,3,4].
After adding 1 to A[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
After adding -3 to A[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
After adding -4 to A[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
After adding 2 to A[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.

Note:

1. 1 <= A.length <= 10000
2. -10000 <= A[i] <= 10000
3. 1 <= queries.length <= 10000
4. -10000 <= queries[i][0] <= 10000
5. 0 <= queries[i][1] < A.length

Solution(TLC):

class Solution {
public:
    vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries) 
    {
        vector<int> vec;vec.clear();
        for (auto row : queries)
        {
            if(row.size() == 2)
            {
                A[row[1]]+=row[0];
            }
            int result = 0;
            for(auto item : A)
            {
                if(item % 2 == 0)
                {
                  result +=item;
                }
            }
            vec.push_back(result);
        }
        return vec;
        
    }
};

原本以为可以EZ水过的，结果成功超时，不出所料，简单题也有坑，算法的时间复杂度过高，看样子需要想一个简单算法,然后能力不足的我颤抖着打开了solution...

Solution(AC):

class Solution {
public:
    vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries)     
{
       vector<int> vec;vec.clear();
        int result = 0;
        for(int i : A)
        {
            if(i%2==0)
            {
                result+=i;
            }
        }
        for(auto row : queries)
        {
            int val = row[0];
            int index = row[1];
            if(A[index] % 2 == 0)
            result -=A[index];
            A[index] +=val;
            if(A[index]%2 == 0)
            result+=A[index];
            
            vec.push_back(result);
        }
        return vec;
        
    }

思路解析：之前的暴力算法没必要每次查询就全部重新算一次，其实每次需要改变的就一个数字，也就是所谓的“增量更新”。

首先把原始所有的偶数全部求和，然后对查询位置上的数奇偶进行判断，如果是偶数则减去（后面会把查询后的结果返回），然后判断查询结果的奇偶，如果是偶数，则加到最终结果。

果然，偷瞄一时爽，一直偷瞄一直爽！感觉题目好坑。。但是这才是ez的题啊。