Leetcode 888. Fair Candy Swap

题目

链接：https://leetcode.com/problems/fair-candy-swap/

Level: Easy

Discription:
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]

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.

代码

class Solution {
public:
    vector fairCandySwap(vector& A, vector& B) {
         int Asum=0,Bsum=0;
         bitset<200002> bit;
         for(auto n:A)
             Asum+=n;
         for(auto n:B)
         {
             Bsum+=n;
             bit.set(n);
         }
         int diff=(Asum-Bsum)/2;
        
         for(int i=0;i0 && bit.test(A[i]-diff))  
            {
                return {A[i], A[i]-diff};
            }
         }
    }
};

思考

• 算法时间复杂度为O(m+n),空间复杂度为O(n)，m为A数组长度，n为B数组长度。
• 如果不用哈希表标记，时间复杂度就是O(mn)。此外用bitset来标记可以节省32倍的空间，并且也可以节约时间。
• 因为返回的数组只有两个值，因此不需要再新建一个vector，直接return {A[i], A[i]-diff}即可。

转载于:https://www.cnblogs.com/zuotongbin/p/10219616.html