• LeetCode每日一题(985. Sum of Even Numbers After Queries)


    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:

    • 1 <= nums.length <= 104
    • -104 <= nums[i] <= 104
    • 1 <= queries.length <= 104
    • -104 <= vali <= 104
    • 0 <= indexi < nums.length

    考虑 4 种情况:

    1. nums[i]原来是偶数, 加上 val 之后还是偶数, 这样偶数的和增加了 val
    2. nums[i]原来是偶数, 加上 val 之后变成奇数, 这样偶数的和减少了 nums[i]
    3. nums[i]原来是奇数, 加上 val 之后变成偶数, 这样偶数的和增加了 nums[i] + val
    4. nums[i]原来是奇数, 加上 val 之后还是奇数, 这样偶数的和没有任何变化

    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
        }
    }
    
    
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  • 原文地址:https://blog.csdn.net/wangjun861205/article/details/125447241