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 <= A.length <= 10000-10000 <= A[i] <= 100001 <= queries.length <= 10000-10000 <= queries[i][0] <= 100000 <= queries[i][1] < A.length
这道题给了一个数组A，说每次可以给数组某个位置上的数字加上一个值，每次操作后让返回当前数组中的偶数数之和。通过题目中的例子可以发现，加上的数字可能为负值，负偶数也是偶数。每次修改一个值后都要返回偶数之和，肯定不能每次都遍历一遍数组求偶数和，太不高效了，其实每次只修改了一个数字，这个数字对整个数组的偶数和的影响有限，可以分情况来讨论一下。假如修改之前，该数字就是偶数，修改后若变为奇数，则损失了原来的偶数值，若修改后还是偶数，则相当于先损失了原来的偶数，又加上了新的偶数。若修改之前，该数字是奇数，修改后若还是奇数，则什么也不影响，若修改后变为了偶数，则相当于加上了这个偶数。所以归纳起来就是，先判断修改前的数字，若是偶数，则减去这个偶数，再判断修改后的数字，若是偶数，则加上这个偶数。这样的话只要最开始遍历一遍数组，求出所有偶数之和，之后修改数字时只要按上面的步骤就可以快速获得偶数之和了，参见代码如下：
class Solution {
public:
vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries) {
vector<int> res;
int n = A.size(), even = 0;
for (int num : A) {
if (num % 2 == 0) even += num;
}
for (auto &query : queries) {
int old = A[query[1]], cur = old + query[0];
if (old % 2 == 0) even -= old;
if (cur % 2 == 0) even += cur;
A[query[1]] = cur;
res.push_back(even);
}
return res;
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/985
参考资料：
https://leetcode.com/problems/sum-of-even-numbers-after-queries/