leetcode 888. Fair Candy Swap 题解【Weekly Content 98 I】

本文介绍了一个名为FairCandySwap的问题解决方法，该问题要求通过一次糖果交换使两个朋友拥有的糖果总量相等。核心思想是计算两个朋友糖果总量之差的一半，并在其中一个集合中寻找与差值匹配的糖果大小，实现高效交换。

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 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.

题解：

核心思想是求出A的和sumA，B的和sumB，两者的差dis == (sumA - sumB) / 2。只要在B中找到一个数B[j]，有B[j] + dis为A中的某个数A[i]，即可交换。如Example1中，sumA为2，sumB为4，则有dis = -1，B[0] + dis = 1，A[0]为1，即为所求。
原理：
A[0] + A[1] + ... + A[i] + ... + A[na-1] = sumA = sumB + 2 * dis
B[0] + B[1] + ... + B[j] + ... + B[nb-1] = sumB
若A[i] == B[j] + dis，则：
A[0] + A[1] + ... + B[j] + dis + ... + A[na-1] = sumA = sumB + 2 * dis
A[0] + A[1] + ... + B[j] + ... + A[na-1] = sumB + dis (1)

B[0] + B[1] + ... + (B[j] + dis) ... + B[nb-1] = sumB + dis
B[0] + B[1] + ... + A[i] ... + B[nb-1] = sumB + dis (2)
易知：
A[0] + A[1] + ... + B[j] + ... + A[na-1] = B[0] + B[1] + ... + A[i] ... + B[nb-1]

C++

class Solution {
public:
    vector<int> fairCandySwap(vector<int>& A, vector<int>& B) {
        int sumA = 0, sumB = 0;
        int lenB = B.size();
        for(auto a: A) {
            sumA += a;
        }
        for(auto b : B) {
            sumB += b;
        }
        int dis = (sumA - sumB) /2;
        unordered_set<int> setA(A.begin(), A.end()); //将A转为set，提升查找效率
        for(int i = 0; i < lenB; i++) {
            if(setA.count(dis + B[i])) {
                return {dis + B[i], B[i]};
            }
        }
        return {};
    }
};