You are given an integer array nums and an array queries where queries[i] = [vali, indexi].
For each query i, first, apply nums[indexi] = nums[indexi] + vali, then print the sum of the even values of nums.
Return an integer array answer where answer[i] is the answer to the ith query.
Example 1:
Input: nums = [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 nums[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
After adding -3 to nums[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
After adding -4 to nums[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
After adding 2 to nums[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.
Example 2:
Input: nums = [1], queries = [[4,0]]
Output: [0]
Constraints:
考虑 4 种情况:
impl Solution {
pub fn sum_even_after_queries(mut nums: Vec<i32>, queries: Vec<Vec<i32>>) -> Vec<i32> {
let mut sum = nums.iter().filter(|v| *v % 2 == 0).map(|v| *v).sum();
let mut ans = Vec::new();
for q in queries {
let curr = nums[q[1] as usize];
if curr % 2 == 0 {
if (curr + q[0]) % 2 == 0 {
sum += q[0];
} else {
sum -= curr;
}
} else {
if (curr + q[0]) % 2 == 0 {
sum += curr + q[0];
}
}
nums[q[1] as usize] += q[0];
ans.push(sum);
}
ans
}
}