README.md

888. 公平的糖果棒交换

English Version

题目描述

爱丽丝和鲍勃有不同大小的糖果棒：A[i] 是爱丽丝拥有的第 i 根糖果棒的大小，B[j] 是鲍勃拥有的第 j 根糖果棒的大小。

因为他们是朋友，所以他们想交换一根糖果棒，这样交换后，他们都有相同的糖果总量。（一个人拥有的糖果总量是他们拥有的糖果棒大小的总和。）

返回一个整数数组 ans，其中 ans[0] 是爱丽丝必须交换的糖果棒的大小，ans[1] 是 Bob 必须交换的糖果棒的大小。

如果有多个答案，你可以返回其中任何一个。保证答案存在。

 

示例 1：

输入：A = [1,1], B = [2,2]
输出：[1,2]

示例 2：

输入：A = [1,2], B = [2,3]
输出：[1,2]

示例 3：

输入：A = [2], B = [1,3]
输出：[2,3]

示例 4：

输入：A = [1,2,5], B = [2,4]
输出：[5,4]

 

提示：

• 1 <= A.length <= 10000
• 1 <= B.length <= 10000
• 1 <= A[i] <= 100000
• 1 <= B[i] <= 100000
• 保证爱丽丝与鲍勃的糖果总量不同。
• 答案肯定存在。

解法

哈希表实现。

Python3

class Solution:
    def fairCandySwap(self, aliceSizes: List[int], bobSizes: List[int]) -> List[int]:
        diff = (sum(aliceSizes) - sum(bobSizes)) >> 1
        s = set(bobSizes)
        for a in aliceSizes:
            target = a - diff
            if target in s:
                return [a, target]

Java

class Solution {
    public int[] fairCandySwap(int[] aliceSizes, int[] bobSizes) {
        int s1 = 0, s2 = 0;
        Set<Integer> s = new HashSet<>();
        for (int a : aliceSizes) {
            s1 += a;
        }
        for (int b : bobSizes) {
            s.add(b);
            s2 += b;
        }
        int diff = (s1 - s2) >> 1;
        for (int a : aliceSizes) {
            int target = a - diff;
            if (s.contains(target)) {
                return new int[]{a, target};
            }
        }
        return null;
    }
}

TypeScript

function fairCandySwap(aliceSizes: number[], bobSizes: number[]): number[] {
    let s1 = aliceSizes.reduce((a, c) => a + c, 0);
    let s2 = bobSizes.reduce((a, c) => a + c, 0);
    let diff = (s1 - s2) >> 1;
    for (let num of aliceSizes) {
        let target = num - diff;
        if (bobSizes.includes(target)) {
            return [num, target];
        }
    }
};

C++

class Solution {
public:
    vector<int> fairCandySwap(vector<int>& aliceSizes, vector<int>& bobSizes) {
        int s1 = accumulate(aliceSizes.begin(), aliceSizes.end(), 0);
        int s2 = accumulate(bobSizes.begin(), bobSizes.end(), 0);
        int diff = (s1 - s2) >> 1;
        unordered_set<int> s(bobSizes.begin(), bobSizes.end());
        vector<int> ans;
        for (int& a : aliceSizes) {
            int target = a - diff;
            if (s.count(target)) {
                ans = vector<int>{a, target};
                break;
            }
        }
        return ans;
    }
};

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