给定一个数组arr []由N 0 s(从1开始的索引)和另一个数组query []组成，每行的格式为{L，R} ，每个查询(L，R)的任务是添加一个(i – L + 1)的值在[L，R]范围内，并打印执行所有查询后获得的数组arr [] 。
例子：
Input: arr[] = {0, 0, 0}, query[][] = {{1, 3}, {2, 3}}
Output: 1 3 5
Explanation: Initially the array is {0, 0, 0}.
Query 1: Range of indices involved: [1, 3]. The value (i – 1 + 1) for each index i in the range is {1, 2, 3}. Adding these values modifies the array to {1, 2, 3}.
Query 2: Range of indices involved: [2, 3]. The value (i – 2 + 1) for each index i in the range is {0, 1, 2}. Adding these values modifies the array to {1, 3, 5}.
Therefore, the modified array is {1, 3, 5}.
Input: arr[] = {0, 0, 0, 0, 0, 0, 0}, query[][] = {{1, 7}, {3, 6}, {4, 5}}
Output: 1 2 4 7 10 10 7
天真的方法：解决给定问题的最简单方法是遍历[L，R]范围内的给定数组，并为每个查询将值(i – L + 1)添加到范围内的每个元素。完成所有查询后，打印获得的修改后的数组arr [] 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to perform the given queries
// in the given empty array of size N
void updateQuery(vector > queries,
int N)
{
// Initialize an array a[]
vector a(N + 1, 0);
// Stores the size of the array
int n = N + 1;
int q = queries.size();
// Traverse the queries
for (int i = 0; i < q; i++) {
// Starting index
int l = queries[i][0];
// Ending index
int r = queries[i][1];
// Increment each index from L to
// R in a[] by (j - l + 1)
for (int j = l; j <= r; j++) {
a[j] += (j - l + 1);
}
}
// Print the modified array
for (int i = 1; i <= N; i++) {
cout << a[i] << " ";
}
}
// Driver Code
int main()
{
int N = 7;
vector > queries
= { { 1, 7 }, { 3, 6 }, { 4, 5 } };
// Function Call
updateQuery(queries, N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to perform the given queries
// in the given empty array of size N
static void updateQuery(int [][]queries, int N)
{
// Initialize an array a[]
ArrayList a = new ArrayList();
for(int i = 0; i < N + 1; i++)
a.add(0);
// Stores the size of the array
int q = 3;
// Traverse the queries
for(int i = 0; i < q; i++)
{
// Starting index
int l = queries[i][0];
// Ending index
int r = queries[i][1];
// Increment each index from L to
// R in a[] by (j - l + 1)
for(int j = l; j <= r; j++)
{
a.set(j, a.get(j)+(j - l + 1));
}
}
// Print the modified array
for(int i = 1; i < a.size(); i++)
{
System.out.print(a.get(i) + " ");
}
}
// Driver code
public static void main(String[] args)
{
int N = 7;
int[][] queries = { { 1, 7 },
{ 3, 6 },
{ 4, 5 } };
// Function Call
updateQuery(queries, N);
}
}
// This code is contributed by offbeat Python3
# Python 3 program for the above approach
# Function to perform the given queries
# in the given empty array of size N
def updateQuery(queries, N):
# Initialize an array a[]
a = [0 for i in range(N + 1)]
# Stores the size of the array
n = N + 1
q = len(queries)
# Traverse the queries
for i in range(q):
# Starting index
l = queries[i][0]
# Ending index
r = queries[i][1]
# Increment each index from L to
# R in a[] by (j - l + 1)
for j in range(l, r + 1, 1):
a[j] += (j - l + 1)
# Print the modified array
for i in range(1, N + 1, 1):
print(a[i], end = " ")
# Driver Code
if __name__ == '__main__':
N = 7
queries = [[1, 7],[3, 6],[4, 5]]
# Function Call
updateQuery(queries, N)
# This code is contributed by ipg2016107.C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to perform the given queries
// in the given empty array of size N
static void updateQuery(int [,]queries, int N)
{
// Initialize an array a[]
List a = new List();
for(int i = 0; i < N + 1; i++)
a.Add(0);
// Stores the size of the array
int q = 3;
// Traverse the queries
for(int i = 0; i < q; i++)
{
// Starting index
int l = queries[i, 0];
// Ending index
int r = queries[i, 1];
// Increment each index from L to
// R in a[] by (j - l + 1)
for(int j = l; j <= r; j++)
{
a[j] += (j - l + 1);
}
}
// Print the modified array
for(int i = 1; i < a.Count; i++)
{
Console.Write(a[i] + " ");
}
}
// Driver Code
public static void Main()
{
int N = 7;
int[,] queries = new int[3, 2] { { 1, 7 },
{ 3, 6 },
{ 4, 5 } };
// Function Call
updateQuery(queries, N);
}
}
// This code is contributed by SURENDRA_GANGWAR C++
// C++ program for the above approach
#include
using namespace std;
// Function to perform the given queries
// in the given empty array of size N
void updateQuery(vector > queries,
int N)
{
// Stores the resultant array
// and the difference array
vector ans(N + 2, 0),
res(N + 2, 0);
int q = queries.size();
// Traverse the given queries
for (int i = 0; i < q; i++) {
// Starting index
int l = queries[i][0];
// Ending index
int r = queries[i][1];
// Increment l-th index by 1
ans[l]++;
// Decrease r-th index by 1
ans[r + 1]--;
// Decrease (r + 1)th index by
// the length of each query
res[r + 1] -= (r - l + 1);
}
// Find the prefix sum of ans[]
for (int i = 1; i <= N; i++)
ans[i] += ans[i - 1];
// Find the final array
for (int i = 1; i <= N; i++)
res[i] += res[i - 1] + ans[i];
// Printing the modified array
for (int i = 1; i <= N; i++) {
cout << res[i] << " ";
}
cout << "\n";
}
// Driver Code
int main()
{
int N = 7;
vector > queries
= { { 1, 7 }, { 3, 6 }, { 4, 5 } };
updateQuery(queries, N);
return 0;
} 输出：
1 2 4 7 10 10 7时间复杂度： O(N * Q)
辅助空间： O(1)
高效的方法：可以通过使用“前缀总和”来优化上述方法。请按照以下步骤解决此问题：
- 使用所有元素初始化为0的数组ans []来存储当前索引受查询影响的次数。
- 使用所有元素都为0初始化数组res [] ，以存储在达到某个查询范围的末尾后要删除的值。
- 遍历给定的查询数组query []并执行以下步骤：
- 将1加到ans [query [i] [0]]并从ans [query [i] [1] + 1]中减去1 。
- 从res [query [i] [1] +1]中减去(query [i] [1] – query [i] [0] +1) 。
- 找到数组ans []的前缀和。
- 遍历数组res []并将每个元素res [i]更新为res [i] + res [i – 1] + ans [i] 。
- 完成上述步骤后，在执行给定查询后，将数组res []打印为修改后的数组。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to perform the given queries
// in the given empty array of size N
void updateQuery(vector > queries,
int N)
{
// Stores the resultant array
// and the difference array
vector ans(N + 2, 0),
res(N + 2, 0);
int q = queries.size();
// Traverse the given queries
for (int i = 0; i < q; i++) {
// Starting index
int l = queries[i][0];
// Ending index
int r = queries[i][1];
// Increment l-th index by 1
ans[l]++;
// Decrease r-th index by 1
ans[r + 1]--;
// Decrease (r + 1)th index by
// the length of each query
res[r + 1] -= (r - l + 1);
}
// Find the prefix sum of ans[]
for (int i = 1; i <= N; i++)
ans[i] += ans[i - 1];
// Find the final array
for (int i = 1; i <= N; i++)
res[i] += res[i - 1] + ans[i];
// Printing the modified array
for (int i = 1; i <= N; i++) {
cout << res[i] << " ";
}
cout << "\n";
}
// Driver Code
int main()
{
int N = 7;
vector > queries
= { { 1, 7 }, { 3, 6 }, { 4, 5 } };
updateQuery(queries, N);
return 0;
}
输出：
1 2 4 7 10 10 7时间复杂度： O(N)
辅助空间： O(N)