【LEETCODE】47、985. Sum of Even Numbers After Queries

package y2019.Algorithm.array;

/**
 * @ProjectName: cutter-point
 * @Package: y2019.Algorithm.array
 * @ClassName: SumEvenAfterQueries
 * @Author: xiaof
 * @Description: 985. Sum of Even Numbers After Queries
 *
 * 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.
 *
 * 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.
 *
 * @Date: 2019/7/4 16:35
 * @Version: 1.0
 */
public class SumEvenAfterQueries {

    public int[] solution(int[] A, int[][] queries) {

        int[] result = new int[queries.length];
        for(int i = 0; i < queries.length; ++i) {
            //计算操作
            int[] temp = queries[i];
            A[temp[1]] = A[temp[1]] + temp[0];
            //计算相应位置的数据
            int tempResult = 0;
            for(int j = 0; j < A.length; ++j) {
                if((A[j] & 1) == 0) {
                    tempResult += A[j];
                }
            }
            result[i] = tempResult;
        }

        return result;
    }

    public static void main(String args[]) {

        int[] A = {1,2,3,4};
        int[][] queries = {{1,0},{-3,1},{-4,0},{2,3}};

        SumEvenAfterQueries fuc = new SumEvenAfterQueries();
        System.out.println(fuc.solution(A, queries));
    }
}