题目地址:https://leetcode.com/problems/sum-of-even-numbers-after-queries/

题目描述

Wehave an array A of integers, and an array queries of queries.

Forthe 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、 1<=A.length<=10000;
2、 -10000<=A[i]<=10000;
3、 1<=queries.length<=10000;
4、 -10000<=queries[i][0]<=10000;
5、 0<=queries[iopeninnewwindow][]<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queriesiopen in new window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;
    }
};

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找规律

上面的暴力解法显然不够优美。根据@votrubac的解法open in new window,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;
    }
};

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