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985. 查询后的偶数和

English Version

题目描述

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

 

提示：

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

解法

Python3

class Solution:
    def sumEvenAfterQueries(
        self, nums: List[int], queries: List[List[int]]
    ) -> List[int]:
        ans = []
        s = sum(num for num in nums if num % 2 == 0)
        for v, i in queries:
            old = nums[i]
            nums[i] += v
            if nums[i] % 2 == 0 and old % 2 == 0:
                s += v
            elif nums[i] % 2 == 0 and old % 2 == 1:
                s += nums[i]
            elif old % 2 == 0:
                s -= old
            ans.append(s)
        return ans

Java

class Solution {
    public int[] sumEvenAfterQueries(int[] nums, int[][] queries) {
        int s = 0;
        for (int num : nums) {
            if (num % 2 == 0) {
                s += num;
            }
        }
        int[] ans = new int[queries.length];
        int idx = 0;
        for (int[] q : queries) {
            int v = q[0], i = q[1];
            int old = nums[i];
            nums[i] += v;
            if (nums[i] % 2 == 0 && old % 2 == 0) {
                s += v;
            } else if (nums[i] % 2 == 0 && old % 2 != 0) {
                s += nums[i];
            } else if (old % 2 == 0) {
                s -= old;
            }
            ans[idx++] = s;
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> sumEvenAfterQueries(vector<int>& nums, vector<vector<int>>& queries) {
        int s = 0;
        for (int& num : nums)
            if (num % 2 == 0)
                s += num;
        vector<int> ans;
        for (auto& q : queries) {
            int v = q[0], i = q[1];
            int old = nums[i];
            nums[i] += v;
            if (nums[i] % 2 == 0 && old % 2 == 0)
                s += v;
            else if (nums[i] % 2 == 0 && old % 2 != 0)
                s += nums[i];
            else if (old % 2 == 0)
                s -= old;
            ans.push_back(s);
        }
        return ans;
    }
};

Go

func sumEvenAfterQueries(nums []int, queries [][]int) []int {
	s := 0
	for _, num := range nums {
		if num%2 == 0 {
			s += num
		}
	}
	var ans []int
	for _, q := range queries {
		v, i := q[0], q[1]
		old := nums[i]
		nums[i] += v
		if nums[i]%2 == 0 && old%2 == 0 {
			s += v
		} else if nums[i]%2 == 0 && old%2 != 0 {
			s += nums[i]
		} else if old%2 == 0 {
			s -= old
		}
		ans = append(ans, s)
	}
	return ans
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number[][]} queries
 * @return {number[]}
 */
var sumEvenAfterQueries = function (nums, queries) {
    let s = 0;
    for (let num of nums) {
        if (num % 2 == 0) {
            s += num;
        }
    }
    let ans = [];
    for (let [v, i] of queries) {
        const old = nums[i];
        nums[i] += v;
        if (nums[i] % 2 == 0 && old % 2 == 0) {
            s += v;
        } else if (nums[i] % 2 == 0 && old % 2 != 0) {
            s += nums[i];
        } else if (old % 2 == 0) {
            s -= old;
        }
        ans.push(s);
    }
    return ans;
};

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