[Swift Weekly Contest 122]LeetCode985. 查询后的偶数和 | Sum of Even Numbers After Queries

概述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 o

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,[2,3]] Output: [8,6,4] Explanation: At the beginning,the array is [1,4]. After adding 1 to A[0],the array is [2,and the sum of even values is 2 + 2 + 4 = 8. After adding -3 to A[1],-1,and the sum of even values is 2 + 4 = 6. After adding -4 to A[0],the array is [-2,and the sum of even values is -2 + 4 = 2. After adding 2 to A[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

给出一个整数数组 A 和一个查询数组 querIEs。

对于第 i 次查询，有 val = querIEs[i][0],index = querIEs[i][1]，我们会把 val 加到 A[index] 上。然后，第 i 次查询的答案是 A 中偶数值的和。

（此处给定的 index = querIEs[i][1] 是从 0 开始的索引，每次查询都会永久修改数组 A。）

返回所有查询的答案。你的答案应当以数组 answer 给出，answer[i] 为第 i 次查询的答案。

示例：

输入：A = [1,querIEs = [[1,3]]输出：[8,4]解释：开始时，数组为 [1,4]。将 1 加到 A[0] 上之后，数组为 [2,4]，偶数值之和为 2 + 2 + 4 = 8。将 -3 加到 A[1] 上之后，数组为 [2,4]，偶数值之和为 2 + 4 = 6。将 -4 加到 A[0] 上之后，数组为 [-2,4]，偶数值之和为 -2 + 4 = 2。将 2 加到 A[3] 上之后，数组为 [-2,6]，偶数值之和为 -2 + 6 = 4。

提示：

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

 772 ms

 1 class Solution { 2     func sumEvenAfterQuerIEs(_ A: [Int],_ querIEs: [[Int]]) -> [Int] { 3         var A = A 4         var s:Int = 0 5         for v in A 6         { 7             if v % 2 == 0 8             { 9                 s += v           10             }11         }12         var q:Int = querIEs.count13         var ret:[Int] = [Int](repeating:0,count:q)14         for i in 0..<q15         {16             var d:Int = querIEs[i][0]17             var pos:Int = querIEs[i][1]18             if A[pos] % 2 == 019             {20                 s -= A[pos]21             }22             A[pos] += d23             if A[pos] % 2 == 024             {25                 s += A[pos]26             }27             ret[i] = s28         }29         return ret30     }31 }

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