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

888. Fair Candy Swap
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.

問題:
核心思想はAの和sumA、Bの和sumB、両者の差dis == (sumA - sumB) / 2.Bの中に1つの数があるB[j]、あるB[j] + disはAのうちのいずれかの数A[i]であれば交換可能です.Example1中、sumAは2、sumBは4はdis = -1、B[0] + dis = 1、A[0]は1は、求められる.原理:A[0] + A[1] + ... + A[i] + ... + A[na-1] = sumA = sumB + 2 * disB[0] + B[1] + ... + B[j] + ... + B[nb-1] = sumBもしA[i] == B[j] + disならば:A[0] + A[1] + ... + B[j] + dis + ... + A[na-1] = sumA = sumB + 2 * disA[0] + A[1] + ... + B[j] + ... + A[na-1] = sumB + dis(1)B[0] + B[1] + ... + (B[j] + dis) ... + B[nb-1] = sumB + disB[0] + B[1] + ... + A[i] ... + B[nb-1] = sumB + dis(2)分かりやすい: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()); // A  set，      
        for(int i = 0; i < lenB; i++) {
            if(setA.count(dis + B[i])) {
                return {dis + B[i], B[i]};
            }
        }
        return {};
    }
};