算法<Array Partition I>

最新推荐文章于 2019-09-26 16:16:55 发布
原创 最新推荐文章于 2019-09-26 16:16:55 发布 · 525 阅读 · 0 ·
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#算法

本文介绍了一种针对数组的特殊配对算法，通过将数组中的2N个元素合理配对，实现每对中较小元素之和的最大化。文章详细阐述了算法原理，并给出了一段简洁高效的Java代码实现。

这个题目的要求是给定一个数组，有2N个元素，将其划分为N对（每一对有两个元素），使得每一对中的最小的元素相加的总和最大，例如：

有一个数组：
s=a1+b1+a2+b2+a3+b(3)+…+an+bn;

我们的目标是将数组划分诸如：
(a1,b1),(a2,b2),(a3,b3),….(an,bn)
然后求：
Sm = min(a1, b1) + min(a2, b2) + … + min(an, bn)
假设对于每一组bi>=ai,Sm(Sm = a1 + a2 + … + an)的最大值就是我们的目标.

定义
1、Sa = a1 + b1 + a2 + b2 + … + an + bn，对于输入的数组来说Sa是一个常数；
2、di = |ai - bi|，Sd = d1 + d2 + … + dn
所以我们可以得到Sa=a1 + a1 + d1 + a2 + a2 + d2 + … + an + an + di = 2Sm + Sd => Sm = (Sa - Sd) / 2

因为Sa是一个常量，要使得Sm最大的话，我们就要使得Sd最小。
一个数组中如何求的Sd最小呢，假如这个数组是一个已排序的数组：
(a1<=b1<=a2<=b2,那么Sd=(b1-a1)+(b2-a2),下图形象地说明了这点：
这里写图片描述
代码实现：

  public int arrayPairSum(int[] nums) {
        Arrays.sort(nums);
        int result = 0;
        for (int i = 0; i < nums.length; i += 2) {
            result += nums[i];
        }
        return result;
    }
Vitis HLS 学习笔记--变量存储（指令：ARRAY_PARTITION）
hi94的博客
04-22 3040
ARRAY_PARTITION 指令中非常重要，它用于优化数组的存储和访问。该指令可以将一个大数组分割成多个小数组，以提高并行处理能力和减少访问延迟。
LeetCode 561 Array Partition I（数组划分）
nomasp
05-05 8222
翻译原文Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a1, b1), (a2, b2), …, (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as pos
LeetCode 561. Array Partition I
mine_soul的博客
04-24 2120
题目描述： Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a1, b1), (a2, b2), …, (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as
Array Partition I
秀秀的博客
12-19 341
问题描述Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a1, b1), (a2, b2), …, (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as pos
#include <stdio.h> #include <string.h> #define N 80 int main(){ char array[N]; scanf("%s",array); int i,j,k=0; for (i=0;array[i]!='\0';i++){ int is_duplicate=0; for(j=0;j<k;j++){ if (array[j]==array[i]){ is_duplicate=1; } } if(!is_duplicate){ array[k]=array[i]; k++; } } char temp; for (i=0;array[i]!='\0';i++){ for (j=i+1;array[j]!='\0';j++){ if (array[i]>array[j]){ temp=array[i]; array[i]=array[j]; array[j]=temp; } } } printf("%s",array); return 0; }
最新发布
11-29
if (array[j] == array[i]) { is_duplicate = 1; break; } } if (!is_duplicate) { array[k] = array[i]; k++; } } // 截断字符串 array[k] = '\0'; // 使用快速排序对字符串进行排序 quickSort...
``` package oy.tol.tra; import java.util.function.Predicate; public class Algorithms { public static <T extends Comparable<T>> void reverse(T[] array) { int i = 0; int j = array.length - 1; while (i < j) { T temp = array[i]; array[i] = array[j]; array[j] = temp; i++; j--; } } public static <T extends Comparable<T>> void sort(T[] array) { boolean swapped; for (int i = 0; i < array.length - 1; i++) { swapped = false; for (int j = 0; j < array.length - 1 - i; j++) { if (array[j].compareTo(array[j + 1]) > 0) { T temp = array[j]; array[j] = array[j + 1]; array[j + 1] = temp; swapped = true; } } if (!swapped) break; } } public static <T extends Comparable<T>> int binarySearch(T aValue, T[] fromArray, int fromIndex, int toIndex) { int low = fromIndex; int high = toIndex ; while (low <= high) { int mid = low + (high - low) / 2; int cmp = fromArray[mid].compareTo(aValue); if (cmp < 0) { low = mid + 1; } else if (cmp > 0) { high = mid - 1; } else { return mid; } } return -1; } public static <E extends Comparable<E>> void fastSort(E[] array) { quickSort(array, 0, array.length - 1); } private static <E extends Comparable<E>> void quickSort(E[] array, int begin, int end) { if (begin < end) { int partitionIndex = partition(array, begin, end); quickSort(array, begin, partitionIndex - 1); quickSort(array, partitionIndex + 1, end); } } public static <T> int partition(T[] array, int count, Predicate<T> rule) { if (array == null || count <= 0) { return 0; // No elements to partition } int left = 0; // Pointer for elements that satisfy the rule int right = count - 1; // Pointer for elements that do not satisfy the rule while (left <= right) { // Move the left pointer to the right until an element does not satisfy the rule while (left <= right && rule.test(array[left])) { left++; } // Move the right pointer to the left until an element satisfies the rule while (left <= right && !rule.test(array[right])) { right--; } // Swap elements if pointers have not crossed if (left <= right) { T temp = array[left]; array[left] = array[right]; array[right] = temp; left++; right--; } } // Return the index of the first element that does not satisfy the rule return left; } }```The method partition(T[], int, Predicate<T>) in the type Algorithms is not applicable for the arguments (E[], int, int)Java(67108979)
03-11
private static <E extends Comparable<E>> int partition(E[] array, int begin, int end) { E pivot = array[end]; int i = begin - 1; for (int j = begin; j < end; j++) { if (array[j].compareTo(pivot) ...
#include <bits/stdc++.h> using namespace std; template <class Type> void Swap(Type& a, Type& b) { Type temp = a; a = b; b = temp; } template <class Type> int Partition(Type a[],int p,int r){ int i = p,j = r+1; Type x = a[p]; while(true){ while(a[++i] < x && i < r); while(a[--j] > x); if(i >= j) break; Swap(a[i],a[j]); } a[p] = a[j]; a[j] = x; return j; } template <class Type> void QuickSort (Type a[],int p,int r){ if(p < r){ int q = Partition<Type>(a, p, r); QuickSort<Type>(a,p,q-1); QuickSort<Type>(a,q+1,r); } } int main() { vector<int> nums; int num; cout << "请输入整数数组："; while (cin >> num) { nums.push_back(num); if (cin.get() == '\n') { break; } } int n = nums.size(); int* intArr = new int[n]; for (int i = 0; i < n; i++) { intArr[i] = nums[i]; } cout << "原始整型数组："; for (int i = 0; i < n; i++) { cout << intArr[i] << " "; } cout << endl; QuickSort<int>(intArr, 0, n - 1); cout << "排序后整型数组："; for (int i = 0; i < n; i++) { cout << intArr[i] << " "; } cout << "\n" << endl; return 0; }画出递归算法流程图
10-17
int partition(std::vector<int>& arr, int low, int high) { int pivot = arr[high]; int i = (low - 1); for (int j = low; j <= high - 1; j++) { if (arr[j] < pivot) { i++; std::swap(arr[i], arr[j]);...
#include <stdio.h> #define MAXLEN 50 typedef int KeyType; typedef struct { KeyType key; } elementType; typedef struct { elementType data[MAXLEN+1]; int len; } SeqList; void creat(SeqList *L) { int i; scanf("%d",&L->len); for(i=1;i<=L->len;i++) scanf("%d",&L->data[i].key); } void print(SeqList L) /*细节在此不表*/ void partition(SeqList *L,int low,int high,int *cutpoint); void quickSort(SeqList *L, int low, int high); int main () { SeqList L; int low,high; creat(&L); low=1; high=L.len; quickSort(&L,low,high); print(L); return 0; } /* 请在这里填写答案 */
06-12
#include <stdio.h> // 交换两个数的值 void swap(int *a, int *b) { int temp = *a; *a = *b; *b = temp; } // 划分数组，使得小于pivot的数在左边，大于pivot的数在右边 int partition(int arr[], int low, ...
Array Partition I(leetcode)
Valerie的博客
10-21 376
Array Partition I Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a1, b1)...(an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possible.
[LeetCode]561. Array Partition I (数组分区 1)
whl_program的博客
04-25 6577
561. Array Partition IGiven an array of 2n integers, your task is to group these integers into n pairs of integer, say (a1, b1), (a2, b2), …, (an, bn) which makes sum of min(ai, bi) for all i from 1 to
[leetcode: Python]561. Array Partition I
cyy2learn的博客
05-19 2393
Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a1, b1), (a2, b2), …, (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possibl
561.Array Partition I
就是那个党伟
09-26 211
int cmp1(const void * a,const void * b) { return (*(int*)a-*(int*)b);//a>b 返回正值 } class Solution { public: int arrayPairSum(vector<int>& nums) { /* 冒泡排序...
561. Array Partition I(C语言)
05-10 1393
这道题什么鬼 不应该输出5，[2,3],一组，[1,4]一组么 Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a1, b1), (a2, b2), ..., (an, bn) which makes sum of min(ai, bi
561. Array Partition I（LeetCode）
weixin_30858241的博客
04-24 145
Given an array of2nintegers, your task is to group these integers intonpairs of integer, say (a1, b1), (a2, b2), ..., (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as poss...
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