985. Sum of Even Numbers After Queries

问题：

给定原数组A，和操作数组queries

假设q=queries[i]

那么q的操作为：对A[q[1]]+=q[0]

evensum为A的所有偶数元素之和，求每次操作后evensum的值，所组成的列表res

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 <= A.length <= 10000
-10000 <= A[i] <= 10000
1 <= queries.length <= 10000
-10000 <= queries[i][0] <= 10000
0 <= queries[i][1] < A.length

　　

解法：

先求的原数组的偶数和evensum

接着遍历操作数组queries

判断操作前A[q[1]]

若为偶数，evensum-=A[q[1]]

进行操作：A[q[1]]+=q[0]

再判断操作后的A[q[1]]

若为偶数，evensum+=A[q[1]]

把evensum放入结果数组res。

代码参考：

class Solution {
public:
    vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries) {
        vector<int> res;
        int evensum=0;
        for(int a:A){
            if(!(a%2))evensum+=a;
        }
        for(vector<int> q:queries){
            if(!(A[q[1]]%2)) evensum-=A[q[1]];
            A[q[1]]+=q[0];
            if(!(A[q[1]]%2)) evensum+=A[q[1]];
            res.push_back(evensum);
        }
        return res;
    }
};

这里，也可以判断操作前A[q[1]]和要追加值q[0]的奇偶性

若为偶+偶，则evensum+=q[0]即可

偶+奇，则evensum-=A[q[1]]即可

奇+奇，则evensum+=(A[q[1]]+q[0])

奇+偶，不变化。

代码参考：

class Solution {
public:
    vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries) {
        vector<int> res;
        int evensum=0;
        for(int a:A){
            if(!(a%2))evensum+=a;
        }
        for(vector<int> q:queries){
            if(!(A[q[1]]%2)){//原数字为偶数
                if(!(q[0]%2)){
                    evensum+=q[0];
                }else{
                    evensum-=A[q[1]];
                }
            }else{//原数字为奇数
                if(q[0]%2){
                    evensum+=(A[q[1]]+q[0]);
                }
            }
            res.push_back(evensum);
            A[q[1]]+=q[0];
        }
        return res;
    }
};