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面试题 16.21. 交换和

English Version

题目描述

给定两个整数数组，请交换一对数值（每个数组中取一个数值），使得两个数组所有元素的和相等。

返回一个数组，第一个元素是第一个数组中要交换的元素，第二个元素是第二个数组中要交换的元素。若有多个答案，返回任意一个均可。若无满足条件的数值，返回空数组。

示例:

输入: array1 = [4, 1, 2, 1, 1, 2], array2 = [3, 6, 3, 3]
输出: [1, 3]

示例:

输入: array1 = [1, 2, 3], array2 = [4, 5, 6]
输出: []

提示：

• 1 <= array1.length, array2.length <= 100000

解法

先计算两个数组的差值 diff，若 diff 为奇数，则说明无满足条件的数值，返回空数组。否则，将 array2 转为 set。然后遍历 array1 中的每个数 e，若值 e - diff 在 set 中，则说明找到满足条件的数值对。

Python3

class Solution:
    def findSwapValues(self, array1: List[int], array2: List[int]) -> List[int]:
        diff = sum(array1) - sum(array2)
        if diff & 1: return []
        diff >>= 1
        s = set(array2)
        for e in array1:
            if (e - diff) in s: return [e, e - diff]
        return []

Java

class Solution {
    public int[] findSwapValues(int[] array1, int[] array2) {
        int diff = sum(array1) - sum(array2);
        if ((diff & 1) == 1) {
            return new int[]{};
        }
        diff >>= 1;
        Set<Integer> s = Arrays.stream(array2).boxed().collect(Collectors.toSet());
        for (int e : array1) {
            if (s.contains((e - diff))) {
                return new int[]{e, e - diff};
            }
        }
        return new int[]{};
    }

    private int sum(int[] array) {
        int res = 0;
        for (int e : array) {
            res += e;
        }
        return res;
    }
}s

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