888. Fair Candy Swap

仅供自己学习

 

题目：

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]

 

思路：

主要是找到两者关系。sumA -x+y =sumB+x-y，即A损失的加上获得的=B损失的加上获得的，如果满足该条件则可获得答案。因为sumA和sumB是已知量，故只需要x = y+（sumA+sumB）/2，判断A和B中是否有对应的x和y即可。只需要将A存入一个哈希表，遍历y并在哈希表中搜索是否有满足等式的x，有则输出答案。

 

代码：

 

class Solution {
public:
    vector<int> fairCandySwap(vector<int>& A, vector<int>& B) {
        int sumA=accumulate(A.begin(),A.end(),0);
        int sumB=accumulate(B.begin(),B.end(),0);
        int delta = (sumA-sumB)/2;
        unordered_set<int> sA(A.begin(),A.end());
        vector<int> res;
        for(auto&y:B)
        {
            int x=y+delta;
            if(sA.count(x))
            {
                res = vector<int> {x,y};
                break;
            }
        }
        return res;
    }
    
};