【LeetCode】985. Sum of Even Numbers After Queries 解题报告(C++)
作者: 负雪明烛
id: fuxuemingzhu
个人博客: http://fuxuemingzhu.cn/
题目地址: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 <= queriesi < A.length
题目大意
给出了原始的数组,然后给出了一串查询步骤,每次查询的时候,都会在指定位置queriesi加上queries[i][0],在每次操作完毕之后,把所有数值是偶数的数字求和保存起来。求经过一系列的查询之后,生成的偶数之和序列是多少。
解题方法
暴力
首先,我们可以按照题目描述使用暴力解法。即每次查询之后都去遍历一次,计算偶数之和,保存起来。
设A的长度是M,queries的次数是M,那么时间复杂度是O(N*M),竟然也通过了。C++用时3000ms,代码如下。
class Solution {
public:
vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries) {
vector<int> res;
for (auto q : queries) {
A[q[1]] += q[0];
int sum = 0;
for (int a : A) {
if (a % 2 == 0) {
sum += a;
}
}
res.push_back(sum);
}
return res;
}
};
找规律
上面的暴力解法显然不够优美。根据@votrubac的解法,我们可以先求出所有偶数之和,然后对于每次查询的时候,如果A[index]是偶数,那么就把这个值减去,然后把查询要添加的数值val加到A[index]上。如果加完的结果是偶数的话,需要把该结果加到sum上。
怎么证明?
首先,我们求出了所有偶数的和。
然后每次查询更改一个数字,有四种更改方式:
- 偶 ==> 奇
- 偶 ==> 偶
- 奇 ==> 奇
- 奇 ==> 偶
所以,如果我们要求在查询之后的偶数和,可以在初始化的偶数和的基础上,先减去在查询之前是偶数的(因为这些偶数已经计算到和里面了,即将变化走了),然后查询是当前的数字进行了变化,然后再加上变化之后是偶数的(因为这些偶数是新变化出来的,需要加到偶数和里面)。这样就求得了新的所有偶数的和。
C++代码如下:
class Solution {
public:
vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries) {
vector<int> sums;
int cursum = 0;
for (int a : A) {
if (a % 2 == 0) {
cursum += a;
}
}
for (auto q : queries) {
if (A[q[1]] % 2 == 0) {
cursum -= A[q[1]];
}
A[q[1]] += q[0];
if (A[q[1]] % 2 == 0) {
cursum += A[q[1]];
}
sums.push_back(cursum);
}
return sums;
}
};
日期
2019 年 2 月 19 日 —— 重拾状态
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