个性化阅读
专注于IT技术分析

算法-差异数组:O(1)时间范围更新查询

本文概述

考虑一个整数数组A[]并遵循以下两种查询。

  1. update(l, r, x):将x添加到从A[l]到A[r]的所有值(包括两者)。
  2. printArray():打印当前修改后的数组。

例子 :

Input : A[] { 10, 5, 20, 40 }
        update(0, 1, 10)
        printArray()
        update(1, 3, 20)
        update(2, 2, 30)
        printArray()
Output : 20 15 20 40
         20 35 70 60
Explanation : The query update(0, 1, 10) 
adds 10 to A[0] and A[1]. After update, A[] becomes {20, 15, 20, 40}       
Query update(1, 3, 20) adds 20 to A[1], A[2] and A[3]. After update, A[] becomes
{20, 35, 40, 60}.
Query update(2, 2, 30) adds 30 to A[2]. 
After update, A[] becomes {20, 35, 70, 60}.

一种简单的解决方案要执行以下操作:

  1. update(l, r, x):从l到r运行循环, 并将x添加到从A[l]到A[r]的所有元素中
  2. printArray():仅打印A[]。

以上两个操作的时间复杂度为O(n)

一个有效的解决方案是使用差异数组。

差异数组

给定数组A[i]的D [i]定义为D [i] = A[i] -A[i-1](对于0 <i <N)和D [0] = A[0]基于0的索引。差异数组可用于执行范围更新查询” l r x”, 其中l是左索引, r是右索引, x是要添加的值, 在所有查询之后, 你都可以从中返回原始数组。可以以O(1)复杂度执行更新范围操作的地方。

  1. update(l, r, x):将x添加到D [l]并将其从D [r + 1]中减去, 即, 我们做D [l] + = x, D [r + 1]-= x
  2. printArray():执行A[0] = D [0]并打印。对于其余元素, 执行A[i] = A[i-1] + D [i]并打印它们。

此处更新的时间复杂度提高到O(1)。请注意, printArray()仍需要O(n)时间。

C ++

//C++ code to demonstrate Difference Array
#include <bits/stdc++.h>
using namespace std;
  
//Creates a diff array D[] for A[] and returns
//it after filling initial values.
vector<int> initializeDiffArray(vector<int>& A)
{
     int n = A.size();
  
     //We use one extra space because
     //update(l, r, x) updates D[r+1]
     vector<int> D(n + 1);
  
     D[0] = A[0], D[n] = 0;
     for ( int i = 1; i <n; i++)
         D[i] = A[i] - A[i - 1];
     return D;
}
  
//Does range update
void update(vector<int>& D, int l, int r, int x)
{
     D[l] += x;
     D[r + 1] -= x;
}
  
//Prints updated Array
int printArray(vector<int>& A, vector<int>& D)
{
     for ( int i = 0; i <A.size(); i++) {
         if (i == 0)
             A[i] = D[i];
  
         //Note that A[0] or D[0] decides
         //values of rest of the elements.
         else
             A[i] = D[i] + A[i - 1];
  
         cout <<A[i] <<" " ;
     }
     cout <<endl;
}
  
//Driver Code
int main()
{
     //Array to be updated
     vector<int> A{ 10, 5, 20, 40 };
  
     //Create and fill difference Array
     vector<int> D = initializeDiffArray(A);
  
     //After below update(l, r, x), the
     //elements should become 20, 15, 20, 40
     update(D, 0, 1, 10);
     printArray(A, D);
  
     //After below updates, the
     //array should become 30, 35, 70, 60
     update(D, 1, 3, 20);
     update(D, 2, 2, 30);
     printArray(A, D);
  
     return 0;
}

Java

//Java code to demonstrate Difference Array
class GFG {
      
     //Creates a diff array D[] for A[] and returns
     //it after filling initial values.
     static void initializeDiffArray( int A[], int D[])
     {
          
         int n = A.length;
  
         D[ 0 ] = A[ 0 ];
         D[n] = 0 ;
         for ( int i = 1 ; i <n; i++)
             D[i] = A[i] - A[i - 1 ];
     }
  
     //Does range update
     static void update( int D[], int l, int r, int x)
     {
         D[l] += x;
         D[r + 1 ] -= x;
     }
  
     //Prints updated Array
     static int printArray( int A[], int D[])
     {
         for ( int i = 0 ; i <A.length; i++) {
              
             if (i == 0 )
                 A[i] = D[i];
  
             //Note that A[0] or D[0] decides
             //values of rest of the elements.
             else
                 A[i] = D[i] + A[i - 1 ];
  
             System.out.print(A[i] + " " );
         }
          
         System.out.println();
          
         return 0 ;
     }
      
     //Driver Code
     public static void main(String[] args)
     {
         //Array to be updated
         int A[] = { 10 , 5 , 20 , 40 };
         int n = A.length;
         //Create and fill difference Array
         //We use one extra space because
         //update(l, r, x) updates D[r+1]
         int D[] = new int [n + 1 ];
         initializeDiffArray(A, D);
  
         //After below update(l, r, x), the
         //elements should become 20, 15, 20, 40
         update(D, 0 , 1 , 10 );
         printArray(A, D);
  
         //After below updates, the
         //array should become 30, 35, 70, 60
         update(D, 1 , 3 , 20 );
         update(D, 2 , 2 , 30 );
          
         printArray(A, D);
     }
}
  
//This code is contributed by Anant Agarwal.

Python3

# Python3 code to demonstrate Difference Array
  
# Creates a diff array D[] for A[] and returns
# it after filling initial values.
def initializeDiffArray( A):
     n = len (A)
  
     # We use one extra space because
     # update(l, r, x) updates D[r+1]
     D = [ 0 for i in range ( 0 , n + 1 )]
  
     D[ 0 ] = A[ 0 ]; D[n] = 0
      
     for i in range ( 1 , n ):
         D[i] = A[i] - A[i - 1 ]
     return D
  
  
# Does range update
def update(D, l, r, x):
  
     D[l] + = x
     D[r + 1 ] - = x
  
  
# Prints updated Array
def printArray(A, D):
  
     for i in range ( 0 , len (A)):
         if (i = = 0 ):
             A[i] = D[i]
  
         # Note that A[0] or D[0] decides
         # values of rest of the elements.
         else :
             A[i] = D[i] + A[i - 1 ]
  
         print (A[i], end = " " )
      
     print ("")
  
  
# Driver Code
A = [ 10 , 5 , 20 , 40 ]
  
# Create and fill difference Array
D = initializeDiffArray(A)
  
# After below update(l, r, x), the
# elements should become 20, 15, 20, 40
update(D, 0 , 1 , 10 )
printArray(A, D)
  
# After below updates, the
# array should become 30, 35, 70, 60
update(D, 1 , 3 , 20 )
update(D, 2 , 2 , 30 )
printArray(A, D)
  
# This code is contributed by Gitanjali.

C#

//C# code to demonstrate Difference Array
using System;
  
class GFG {
      
     //Creates a diff array D[] for A[] and returns
     //it after filling initial values.
     static void initializeDiffArray( int []A, int []D)
     {
          
         int n = A.Length;
  
         D[0] = A[0];
         D[n] = 0;
         for ( int i = 1; i <n; i++)
             D[i] = A[i] - A[i - 1];
     }
  
     //Does range update
     static void update( int []D, int l, int r, int x)
     {
         D[l] += x;
         D[r + 1] -= x;
     }
  
     //Prints updated Array
     static int printArray( int []A, int []D)
     {
         for ( int i = 0; i <A.Length; i++) {
              
             if (i == 0)
                 A[i] = D[i];
  
             //Note that A[0] or D[0] decides
             //values of rest of the elements.
             else
                 A[i] = D[i] + A[i - 1];
  
             Console.Write(A[i] + " " );
         }
          
         Console.WriteLine();
          
         return 0;
     }
      
     //Driver Code
     public static void Main()
     {
         //Array to be updated
         int []A = { 10, 5, 20, 40 };
         int n = A.Length;
         //Create and fill difference Array
         //We use one extra space because
         //update(l, r, x) updates D[r+1]
         int []D = new int [n + 1];
         initializeDiffArray(A, D);
  
         //After below update(l, r, x), the
         //elements should become 20, 15, 20, 40
         update(D, 0, 1, 10);
         printArray(A, D);
  
         //After below updates, the
         //array should become 30, 35, 70, 60
         update(D, 1, 3, 20);
         update(D, 2, 2, 30);
          
         printArray(A, D);
     }
}
  
//This code is contributed by vt_m.

输出如下:

20 15 20 40 
20 35 70 60

赞(0) 打赏
未经允许不得转载:srcmini » 算法-差异数组:O(1)时间范围更新查询
分享到: 更多 (0)

评论 抢沙发

评论前必须登录!

 

觉得文章有用就打赏一下文章作者

微信扫一扫打赏