985. Sum of Even Numbers After Queries*
https://leetcode.com/problems/sum-of-even-numbers-after-queries/
题目描述
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] <= 10000 -
1 <= queries.length <= 10000 -
-10000 <= queries[i][0] <= 10000 -
0 <= queries[i][1] < A.length
C++ 实现 1
思路很暴力, 使用 odd_sum 始终维护数组的奇数和, 使用 even_sum 始终维护数组的偶数和.
class Solution {
public:
vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries) {
int odd_sum = 0, even_sum = 0;
for (auto &a : A) {
if (a % 2 == 0) even_sum += a;
else odd_sum += a;
}
vector<int> res;
for (auto &q : queries) {
int val = q[0], index = q[1];
if (A[index] % 2 == 0) {
if (val % 2 == 0) {
A[index] += val;
even_sum += val;
} else {
even_sum -= A[index];
A[index] += val;
odd_sum += A[index];
}
} else {
if (val % 2 == 0) {
A[index] += val;
odd_sum += val;
} else {
odd_sum -= A[index];
A[index] += val;
even_sum += A[index];
}
}
res.push_back(even_sum);
}
return res;
}
};
扩展阅读
[Java] 10 liner - odd / even analysis, time O(max(m, n))
