leetcode 888. Fair Candy Swap

最新推荐文章于 2024-08-17 10:58:13 发布

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最新推荐文章于 2024-08-17 10:58:13 发布

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博客围绕Alice和Bob交换糖果展开，目的是交换后两人糖果总量相同。给出了从两序列中各选两数交换使序列和相同的问题，还给出计算公式，若记sumA、sumB分别为两序列和，dif为 (sumA - sumB)/2 ，则有 B[] = A[] - dif。

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]

从所给的两序列中，各选两个数交换，交换后的两个序列的和相同。
记 sumA为vector A的和，sumB为vector B的和，A[] 为A容器中的某个数字，B[] 为B容器中的某个数，则：
sumA-2*A[]==sumB-2*B[] => B[]=A[]+(sumB-sumA)/2
若记dif 为 (sumA-sumB)/2,有 B[]=A[]-dif

#include <algorithm>
using namespace std;
class Solution {
public:
    vector<int> fairCandySwap(vector<int>& A, vector<int>& B) {
        vector<bool> exist(100005,false); 
        vector<int> ans;
        int sumA=0;
        bool symbol=true;
        for(vector<int>::iterator pos=A.begin();pos<A.end();pos++)
            sumA+=*pos;
        int sumB=0;
        for(vector<int>::iterator pos=B.begin();pos<B.end();pos++)
            sumB+=*pos;
        if(sumA<sumB){
            A.swap(B);
            swap(sumA,sumB);
            symbol=false;
        }
        for(auto pos=B.begin();pos<B.end();pos++)
            exist[*pos]=true;
        int dif=(sumA-sumB)/2;
        for(register int i=0;i<A.size();i++){
            int num=A[i]-dif;
            if(num<=0||num>100000)
                continue;
            if(exist[num]){
                if(symbol){
                    ans.push_back(A[i]);
                    ans.push_back(num);
                }else{
                    ans.push_back(num);
                    ans.push_back(A[i]);
                }
                break;
            }
        }
        return ans;
    }
};