[GeeksForGeeks] Find the largest pair sum in an unsorted array

最新推荐文章于 2025-12-21 21:50:00 发布

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原文链接：http://www.cnblogs.com/lz87/p/7220208.html

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#runtime #java

本文介绍了三种算法来解决寻找未排序数组中最大的两个元素之和的问题。第一种使用排序，复杂度为O(n*logn)；第二种利用堆，实现O(n)时间复杂度；第三种扫描数组一次更新最大及次大的值，同样达到O(n)复杂度。

Given an unsorted array, find the largest pair sum.

Solution 1. O(n*logn) runtime, using sorting

Solution 2. O(n) runtime, using heapify (max priority queue)

Option 1. Use Java priority queue API, easy to implement but uses O(n) extra memory.

Option 2. Since we are given an array, heapify the array to a max pq. This does not use extra memory.

Heapify takes O(n) runtime;

delete the max value from the heapified array takes O(logn) runtime as we need to percolate down the

new head element.

 1 public class Solution {
 2     public int findLargestPair(int[] A){
 3         if(A == null || A.length <= 1){
 4             return Integer.MIN_VALUE;
 5         }
 6         heapify(A);
 7         int firstMax = A[0], secondMax = 0;
 8         A[0] = A[A.length - 1];
 9         percolateDown(A, A.length - 1, 0);
10         secondMax = A[0];
11         return firstMax + secondMax;
12     }
13     private void heapify(int[] A){
14         for(int i = A.length / 2 - 1; i >= 0; i--){
15             percolateDown(A, A.length, i);
16         }
17     }
18     private void percolateDown(int[] A, int len, int idx){
19         int maxIdx = idx;
20         int leftChildIdx = 2 * idx + 1;
21         int rightChildIdx = 2 * idx + 2;
22         if(leftChildIdx < len && A[leftChildIdx] > A[maxIdx]){
23             maxIdx = leftChildIdx;
24         }
25         if(rightChildIdx < len && A[rightChildIdx] > A[maxIdx]){
26             maxIdx = rightChildIdx;
27         }
28         if(maxIdx != idx){
29             int temp = A[idx];
30             A[idx] = A[maxIdx];
31             A[maxIdx] = temp;
32             percolateDown(A, len, maxIdx);
33         }
34     }
35 }

Solution 3. O(n) runtime, O(1) space, scan the input array once and upate the largest and second largest values along the way.

 1 public int findLargestPair(int[] A){
 2     if(A == null || A.length <= 1){
 3         return Integer.MIN_VALUE;
 4     }
 5     int firstMax = Math.max(A[0], A[1]);
 6     int secondMax = Math.min(A[0], A[1]);
 7     for(int i = 2; i < A.length; i++){
 8         if(A[i] >= firstMax){
 9             secondMax = firstMax;
10             firstMax = A[i];
11         }
12         else if(A[i] > secondMax){
13             secondMax = A[i];
14         }
15     }
16     return firstMax + secondMax;
17 }