Problem: 面试题 16.21. 交换和

题目描述

思路及解法

1.分别求取array1与array2数组每一个元素的和（sum1与sum2）并同时将array2的元素存入一个set集合中；
2.如果sum1和sum2的和为奇数，则不存在；
3.计算sum1与sum2的和的一半，同时减去sum1并用变量diff记录下来；
4.遍历array1同时在set集合中查找是否存在（array1[i] + diff）;若存在则交换array1[i]与array1[i] + diff;

复杂度

时间复杂度:

$O(n)$

空间复杂度:

$O(n)$

Code

class Solution {
public:/*** Hash* @param array1 Given array1* @param array2 Given array2* @return vector<int>*/vector<int> findSwapValues(vector<int>& array1, vector<int>& array2) {int len1 = array1.size();int len2 = array2.size();//Calculate the sum of array1int sum1 = 0;for (int i = 0; i < len1; ++i) {sum1 += array1[i];}int sum2 = 0;unordered_set<int> set;//Evaluates the elements of array2 and// stores its elements into the hash collectionfor (int i = 0; i < len2; ++i) {sum2 += array2[i];if (set.find(array2[i]) == set.end()) {set.insert(array2[i]);}}//Does not exist if the sum is oddint sum = sum1 + sum2;if (sum % 2 != 0) {return vector<int>();}//Iterate over each element in array1,// looking it up in the hash tableint diff = sum / 2 - sum1;for (int i = 0; i < len1; ++i) {int target = array1[i] + diff;if (set.find(target) != set.end()) {return vector<int>{array1[i], target};}}return vector<int>();}
};