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、 1 <= A.length <= 10000
2、1 <= B.length <= 10000
3、1 <= A[i] <= 100000
4、1 <= B[i] <= 100000
5、It is guaranteed that Alice and Bob have different total amounts of candy.
6、It is guaranteed there exists an answer.
解决方案：
我能想到的是首先找两个数组的sum中间的数，就可以保证他们可以达到相等，但是sumA和sumB中间可能有许多数，且相加不一定等于sumA+sumB，LeetCode上的大神给出了两种Python解决方案，如下：
Python
（1）
'''
If we know the final same total amount total,
we can pick any element a in A and check if the rest in A + any element b in B == total.
That is, the rest of candy left in A = sum(A) - element in A
possible B = total - (the rest of candy left in A)
'''
class Solution:
def CandySwap(self,A,B):
sumA,sumB = sum(A),sum(B)
setA,setB = set(A),set(B)
total = (sumA+sumB)//2 # "//"表示整除
for i in setA:
item = total-(sumA-i)
if item in setB:
retrun [i,item]
（2）
'''
We know the difference between two arrays.
Lets call it diff.
candy_alice is the candy alice gave to bob
candy_bob is the candy bob gave to alice
Alice's candy after exchange: sumA - candy_alice + candy_bob
Bob's candy after exchange : sumB - candy_bob + candy_alice
These two should equal after exchange:
sumA - candy_alice + candy_bob = sumB - candy_bob + candy_alice
use math we can get: sumA - sumB = 2*(candy_alice - candy_bob) = diff
We can represent candy bob as: candy_alice - diff/2
then, for each element in A, check if candy_alice - diff/2 is in B
'''
class Solution:
def CandySwap(self,A,B):
sumA,sumB = sum(A),sum(B)
setA,setB = set(A),set(B)
diff = sumA - sumB
for i in setA:
item = i - float(diff/2)
if item in setB:
retrun [i,int(item)]