Description
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
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.
Solution
这道题从数学的角度出发比较好理解：设A要分出去x颗糖，B要分出去y颗糖，要使A和B最终拥有相同数量的糖，需要满足sumA - x + y = sumB - y + x，也就是y = x + (sumB - sumA) / 2。
有了以上这个结论后，我们要做的就是先遍历A、B两个数组分别求出sumA和sumB，然后查找A数组中的每个元素（对应着公式中的x）在B数组中是否有对应的元素（对应着公式中的y），如果找到了对应的元素，则这道题就解出了。
class Solution {
public int[] fairCandySwap(int[] A, int[] B) {
int sumA = 0, sumB = 0;
HashSet<Integer> set = new HashSet<>();
for(int x : A)
sumA += x;
for(int x : B) {
sumB += x;
set.add(x);
}
int delta = (sumB - sumA) / 2;
for(int x : A) {
if(set.contains(delta + x))
return new int[] {x, delta+x};
}
return null;
}
}