# 在未排序的数组中找到最大的对和 > 原文： [https://www.geeksforgeeks.org/find-the-largest-pair-sum-in-an-unsorted-array/](https://www.geeksforgeeks.org/find-the-largest-pair-sum-in-an-unsorted-array/) 给定一个未排序的不同整数，在其中找到最大的对和。 例如，`{12, 34, 10, 6, 40}`中最大的对和为 74。 难度等级：新秀 预期时间复杂度：`O(n)`【仅允许遍历数组一次】 这个问题主要归结为找到数组中最大和第二大元素。 通过遍历数组一次，我们可以找到`O(n)`时间中最大和第二大的值。 ``` 1) Initialize both first and second largest first = max(arr[0], arr[1]) second = min(arr[0], arr[1]) 2) Loop through remaining elements (from 3rd to end) a) If the current element is greater than first, then update first and second. b) Else if the current element is greater than second then update second 3) Return (first + second) ``` 下面是上述算法的实现： ## C++ ```cpp // C++ program to find largest pair sum in a given array #include<iostream> using namespace std; /* Function to return largest pair sum. Assumes that     there are at-least  two elements in arr[] */ int findLargestSumPair(int arr[], int n) {     // Initialize first and second largest element     int first, second;     if (arr[0] > arr[1])     {         first = arr[0];         second = arr[1];     }     else     {         first = arr[1];         second = arr[0];     }     // Traverse remaining array and find first and second largest     // elements in overall array     for (int i = 2; i<n; i ++)     {         /* If current element is greater than first then update both           first and second */         if (arr[i] > first)         {             second = first;             first = arr[i];         }         /* If arr[i] is in between first and second then update second  */         else if (arr[i] > second && arr[i] != first)             second = arr[i];     }     return (first + second); } /* Driver program to test above function */ int main() {     int arr[] = {12, 34, 10, 6, 40};     int n = sizeof(arr)/sizeof(arr[0]);     cout << "Max Pair Sum is " << findLargestSumPair(arr, n);     return 0; } ```