在索引 i 可以使 arr[i] 跳转的条件下到达数组末尾的最大跳转

给定一个整数N和一个大小为N的数组arr[ ] ，任务是在给定约束条件下找到到达数组末尾的最大跳跃，即从索引i可以使arr[i]跳跃并到达位置i+arr [我] 。

例子：

Input: N = 5, arr[] = {2, 3, 5, 7, 9}
Output: 12 
Explanation:
At index 0 make 2 jumps and move to index 2 and make 5 jumps after that to reach index 7 which is out of the array so total number of jumps is (2+5)=7. 
At index 1 make 3+9= 12 jumps 
At index 2 make 5 jumps
At index 3 make 7 jumps
At index 4 make 9 jumps 

Input: arr[]={2, 2, 1, 2, 3, 3}
Output: 8

方法：思路是用动态规划来解决这个问题。请按照以下步骤解决问题。

• 用 0 初始化大小为N的数组dp 。 dp[i]存储从索引i到达数组末尾所需的跳转次数。此外，将整数ans初始化为 0。
• 遍历数组 从数组的末尾使用 for 循环
  • 将arr[i]分配给dp[i]，因为arr[i]是从此索引跳转的最小次数。
  • 现在初始化一个变量 say j并赋值j = i+arr[i]。
  • 如果j的值小于N ，则将dp[j]添加到dp[i] 。
  • 将ans的值更新为 max( ans , dp[i] )
• 完成上述步骤后，打印ans的值。

C++

// C++ code for the above approach
#include 
using namespace std;
 
// Function to find the maximum
// jumps to reach end of array
void findMaxJumps(int arr[], int N)
{
    // Stores the jumps needed
    // to reach end from each index
    int dp[N] = { 0 };
    int ans = 0;
 
    // Traverse the array
    for (int i = N - 1; i >= 0; i--) {
        dp[i] = arr[i];
        int j = i + arr[i];
 
        // Check if j is less
        // than N
        if (j < N) {
 
            // Add dp[j] to the
            // value of dp[i]
            dp[i] = dp[i] + dp[j];
        }
 
        // Update the value
        // of ans
        ans = max(ans, dp[i]);
    }
 
    // Print the value of ans
    cout << ans;
}
 
// Driver Code
int main()
{
 
    int arr[] = { 2, 3, 5, 7, 9 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    findMaxJumps(arr, N);
 
    return 0;
}

Java

// Java code for the above approach
 
import java.io.*;
 
class GFG {
   
// Function to find the maximum
// jumps to reach end of array
static void findMaxJumps(int arr[], int N)
{
    // Stores the jumps needed
    // to reach end from each index
    int dp[] = new int [N];
    int ans = 0;
 
    // Traverse the array
    for (int i = N - 1; i >= 0; i--) {
        dp[i] = arr[i];
        int j = i + arr[i];
 
        // Check if j is less
        // than N
        if (j < N) {
 
            // Add dp[j] to the
            // value of dp[i]
            dp[i] = dp[i] + dp[j];
        }
 
        // Update the value
        // of ans
        ans = Math.max(ans, dp[i]);
    }
 
    // Print the value of ans
    System.out.println(ans);
}
 
// Driver Code
public static void main (String[] args) {
    int arr[] = { 2, 3, 5, 7, 9 };
    int N = arr.length;
 
    findMaxJumps(arr, N);
}
}
 
// This code is contributed by Dharanendra L V.

Python3

# python 3 code for the above approach
 
# Function to find the maximum
# jumps to reach end of array
def findMaxJumps(arr, N):
   
    # Stores the jumps needed
    # to reach end from each index
    dp = [0 for i in range(N)]
    ans = 0
 
    # Traverse the array
    i = N - 1
    while(i >= 0):
        dp[i] = arr[i]
        j = i + arr[i]
 
        # Check if j is less
        # than N
        if (j < N):
 
            # Add dp[j] to the
            # value of dp[i]
            dp[i] = dp[i] + dp[j]
 
        # Update the value
        # of ans
        ans = max(ans, dp[i])
        i -= 1
 
    # Print the value of ans
    print(ans)
 
# Driver Code
if __name__ == '__main__':
    arr = [2, 3, 5, 7, 9]
    N =  len(arr)
 
    findMaxJumps(arr, N)
     
    # This code is contributed by ipg2016107.

C#

// C# code for the above approach
 
using System;
 
class GFG {
 
    // Function to find the maximum
    // jumps to reach end of array
    static void findMaxJumps(int[] arr, int N)
    {
        // Stores the jumps needed
        // to reach end from each index
        int[] dp = new int[N];
        int ans = 0;
 
        // Traverse the array
        for (int i = N - 1; i >= 0; i--) {
            dp[i] = arr[i];
            int j = i + arr[i];
 
            // Check if j is less
            // than N
            if (j < N) {
 
                // Add dp[j] to the
                // value of dp[i]
                dp[i] = dp[i] + dp[j];
            }
 
            // Update the value
            // of ans
            ans = Math.Max(ans, dp[i]);
        }
 
        // Print the value of ans
        Console.Write(ans);
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        int[] arr = { 2, 3, 5, 7, 9 };
        int N = arr.Length;
 
        findMaxJumps(arr, N);
    }
}
 
// This code is contributed by ukasp.

Javascript

// Javascript code for the above approach
 
// Function to find the maximum
// jumps to reach end of array
function findMaxJumps(arr, N)
{
 
    // Stores the jumps needed
    // to reach end from each index
    let dp = new Array(N).fill(0);
    let ans = 0;
 
    // Traverse the array
    for (let i = N - 1; i >= 0; i--) {
        dp[i] = arr[i];
        let j = i + arr[i];
 
        // Check if j is less
        // than N
        if (j < N) {
 
            // Add dp[j] to the
            // value of dp[i]
            dp[i] = dp[i] + dp[j];
        }
 
        // Update the value
        // of ans
        ans = Math.max(ans, dp[i]);
    }
 
    // Print the value of ans
    document.write(ans);
}
 
// Driver Code
let arr = [2, 3, 5, 7, 9];
let N = arr.length;
 
findMaxJumps(arr, N);
 
// This code is contributed by _saurabh_jaiswal.

输出

12

时间复杂度： O(N)
辅助空间： O(N)

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