888. Fair Candy Swap

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**888. Fair Candy Swap**

Alice and Bob have candy bars of different sizes: A[i] is the size of the i-th bar of candy that Alice has, and B[j] is the size of the j-th bar of candy that Bob has.

Since they are friends, they would like to exchange one candy bar each so that after the exchange, they both have the same total amount of candy.  (The total amount of candy a person has is the sum of the sizes of candy bars they have.)

Return an integer array ans where ans[0] is the size of the candy bar that Alice must exchange, and ans[1] is the size of the candy bar that Bob must exchange.

If there are multiple answers, you may return any one of them.  It is guaranteed an answer exists.

Example 1:

Input: A = [1,1], B = [2,2]
Output: [1,2]

Example 2:

Input: A = [1,2], B = [2,3]
Output: [1,2]

Example 3:

Input: A = [2], B = [1,3]
Output: [2,3]

Example 4:

Input: A = [1,2,5], B = [2,4]
Output: [5,4]

Note:

• 1 <= A.length <= 10000
• 1 <= B.length <= 10000
• 1 <= A[i] <= 100000
• 1 <= B[i] <= 100000
• It is guaranteed that Alice and Bob have different total amounts of candy.
• It is guaranteed there exists an answer.

题目大意：

有两个数组A、B，从中各自拿出一个数交换，使交换后的两个数组的元素和相等。首先分别计算出A、B的和sumA、sumB，然后计算平均数avg，avg就是A、B交换后的和。

然后依次遍历A，假设遍历到第i个数。
如果把A[i]换掉，那么A中剩余元素的和是：sumA - A[i]。要想使A的各元素和达到avg，还差avg - (sumA - A[i])，记为diff。如果B中刚好有diff，则可以交换diff和A[i]。

class Solution {
    public int[] fairCandySwap(int[] A, int[] B) {        
        int sumA = 0;
        for(int i : A) {
            sumA += i;
        }

        int sumB = 0;
        Set<Integer> set = new HashSet<>(); // 用hashSet提高时间效率
        for(int i : B) {
            set.add(i);
            sumB += i;
        }

        int avg = (sumA + sumB) >> 1; // 计算平均数

        for(int i = 0; i < A.length; i++) {
            // 如果把A[i]换掉，想要使得A的和达到avg，还差diff
            int diff = avg - (sumA - A[i]);
            if(set.contains(diff)) {
                return new int[]{A[i], diff};
            }
        }

        throw null;
    }
}