leetcode-985-Sum-of-Even-Numbers-After-Queries

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:

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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. 1 <= A.length <= 10000
  2. -10000 <= A[i] <= 10000
  3. 1 <= queries.length <= 10000
  4. -10000 <= queries[i][0] <= 10000
  5. 0 <= queries[i][1] < A.length

分析


最简单的方式是直接按照题意两次循环,时间复杂度是 $O(Q*N)$,$Q$ 是 $queries$ 的长度,$N$ 是 $A$ 的长度。空间复杂度是 $O(Q)$。该方法用 Java 实现

Solution 里有一个讨巧的方式,先把 A 的偶数和都计算出来,记为 S,时间复杂度是 $O(N)$,接着遍历 queries,首先判断 A[index] 是否为偶数,如果为偶数,S 要减去 A[index],因为不能把 A[index] 纳入和的计算,接着再根据 A[index] + val 计算真正的和。该方法用 Python 实现。

解决方案1(Java)


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class Solution {
public int[] sumEvenAfterQueries(int[] A, int[][] queries) {
int[] result = new int[A.length];
for (int i = 0; i < queries.length; i++) {
int[] query = queries[i];
int add = query[0];
int index = query[1];
A[index] += add;
int sum = 0;
for (int j = 0; j < A.length; j++) {
if (A[j] % 2 == 0) {
sum += A[j];
}
}
result[i] = sum;
}
return result;
}
}

解决方案1(Python)


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class Solution(object):
def sumEvenAfterQueries(self, A, queries):
"""
:type A: List[int]
:type queries: List[List[int]]
:rtype: List[int]
"""
S = sum(x for x in A if x % 2 == 0)
result = []
for x, k in queries:
if A[k] % 2 == 0:
S -= A[k]
A[k] += x
if A[k] % 2 == 0:
S += A[k]
result.append(S)
return result

题目来源