通过从 arr[i] 跳转到 arr[arr[i]] 到达 Array 末尾的跳转次数

给定一个包含N个整数的数组arr[] ，任务是找出所有i值作为起始索引的数组arr[]转义所需的跳转次数，该索引位于[0, N)范围内，其中唯一可能的跳转来自arr[i]是到arr[arr[i]]并且转义数组意味着arr[i]>=N ，即下一个跳转的索引不存在。

例子：

Input: arr[] = {2, 3, 4, 1, 10}
Output: 3 -1 2 -1 1
Explanation:

For i = 0, initially the current index x is 0. After the 1st jump, the current index becomes x = arr[x] = arr[0] = 2. Similarly, after the 2nd jump, the current index becomes x = arr[2] = 4. After the 3rd jump x = arr[4] =10, which is greater than the array size. Therefore the number of steps required to escape the array is 3.
For i = 1, initially the current index x is 1. After the 1st jump, x = arr[1] = 3. After the 2nd jump, x = arr[3] = 1, which has already been visited and hence forming a closed loop. Therefore it is impossible to escape the array from index 1.

Input: arr[] = {3, 12, 2, 7, 4, 10, 35, 5, 9, 27}
Output: 4 1 -1 3 -1 1 1 2 2 1

编程需要懂一点英语

方法：给定的问题可以在递归的帮助下解决。以下是要遵循的步骤：

创建一个数组visited[] ，它存储当前索引是否已经被访问过。最初， visited = {0} 。
创建一个数组cntJumps[] ，它存储 [0, N) 范围内所有索引所需的跳转次数。最初， cntJumps = {0} 。
创建一个递归函数countJumps() ，它计算从当前索引中转义数组所需的跳转次数。
在函数countJumps()中，如果已经计算出当前索引的答案，则返回 answer else 如果当前节点已经被访问，则返回-1否则如果从当前索引跳转后数组将被转义，则返回1 。
递归计算当前跳转后的跳转次数，即countJumps(i) = 1 + countJumps(arr[i]) 。将答案存储在数组cntJumps[]中。
打印数组cntJumps[] ，这是必需的答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Stores the number of jumps
int cntJumps[100];
 
// Stores if the current index is visited
int visited[100];
 
// Recursive function to find the number of
// jumps to escape the array from index i
int countJumps(int arr[], int N, int i)
{
    // If the answer for the current index is
    // already calculated
    if (cntJumps[i] != 0) {
        return cntJumps[i];
    }
 
    // If the current index is already visited
    if (visited[i]) {
        return -1;
    }
 
    // Mark current index as visited
    visited[i] = true;
 
    // If the array is escaped after the jump
    // from the current index
    if (arr[i] >= N) {
        return cntJumps[i] = 1;
    }
 
    // Recursive call for the next jump
    int val = countJumps(arr, N, arr[i]);
 
    // If it is impossible to escape the array
    if (val == -1)
        cntJumps[i] = -1;
    else
        cntJumps[i] = 1 + val;
 
    // Return answer
    return cntJumps[i];
}
 
// Function to print the number of jumps
// required to escape the array from
// ith index for all values of i in [0, N)
void printCountJumps(int arr[], int N)
{
    // Initialize visited array as 0
    memset(visited, 0, sizeof(visited));
 
    // Initialize cntJump array by 0
    memset(cntJumps, 0, sizeof(cntJumps));
 
    // Loop to iterate over all values of i
    for (int i = 0; i < N; i++) {
 
        // If the index i is not visited already
        if (!visited[i]) {
            countJumps(arr, N, i);
        }
    }
 
    // Print Answer
    for (int i = 0; i < N; i++) {
        cout << cntJumps[i] << " ";
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 12, 2, 7, 4, 10, 35, 5, 9, 27 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    printCountJumps(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
 
public class GFG {
 
// Stores the number of jumps
static int cntJumps[] = new int[100];
 
// Stores if the current index is visited
static int visited[] = new int[100];
 
// Recursive function to find the number of
// jumps to escape the array from index i
static int countJumps(int arr[], int N, int i)
{
    // If the answer for the current index is
    // already calculated
    if (cntJumps[i] != 0) {
        return cntJumps[i];
    }
 
    // If the current index is already visited
    if (visited[i] != 0) {
        return -1;
    }
 
    // Mark current index as visited
    visited[i] = 1;
 
    // If the array is escaped after the jump
    // from the current index
    if (arr[i] >= N) {
        cntJumps[i] = 1;
        return cntJumps[i];
    }
 
    // Recursive call for the next jump
    int val = countJumps(arr, N, arr[i]);
 
    // If it is impossible to escape the array
    if (val == -1)
        cntJumps[i] = -1;
    else
        cntJumps[i] = 1 + val;
 
    // Return answer
    return cntJumps[i];
}
 
// Function to print the number of jumps
// required to escape the array from
// ith index for all values of i in [0, N)
static void printCountJumps(int arr[], int N)
{
    // Initialize visited array as 0
    for (int i = 0; i < visited.length; i++)
        visited[i] = 0;
         
    // Initialize cntJump array by 0
    for (int i = 0; i < cntJumps.length; i++)
        cntJumps[i] = 0;
 
    // Loop to iterate over all values of i
    for (int i = 0; i < N; i++) {
 
        // If the index i is not visited already
        if (visited[i] == 0) {
            countJumps(arr, N, i);
        }
    }
 
    // Print Answer
    for (int i = 0; i < N; i++) {
        System.out.print(cntJumps[i] + " ");
    }
}
 
    // Driver Code
    public static void main (String[] args)
    {
        int arr[] = { 3, 12, 2, 7, 4, 10, 35, 5, 9, 27 };
        int N = arr.length;
     
        printCountJumps(arr, N);
     
    }
}
 
// This code is contributed by AnkThon

Python3

# Python program for the above approach
 
# Stores the number of jumps
cntJumps = [0 for _ in range(100)]
 
# Stores if the current index is visited
visited = [0 for _ in range(100)]
 
# Recursive function to find the number of
# jumps to escape the array from index i
def countJumps(arr, N, i):
    global visited
    global cntJumps
 
    # If the answer for the current index is
    # already calculated
    if (cntJumps[i] != 0):
        return cntJumps[i]
 
        # If the current index is already visited
    if (visited[i]):
        return -1
 
        # Mark current index as visited
    visited[i] = True
 
    # If the array is escaped after the jump
    # from the current index
    if (arr[i] >= N):
        cntJumps[i] = 1
        return cntJumps[i]
 
        # Recursive call for the next jump
    val = countJumps(arr, N, arr[i])
 
    # If it is impossible to escape the array
    if (val == -1):
        cntJumps[i] = -1
    else:
        cntJumps[i] = 1 + val
 
        # Return answer
    return cntJumps[i]
 
 
# Function to print the number of jumps
# required to escape the array from
# ith index for all values of i in [0, N)
def printCountJumps(arr,  N):
 
        # Loop to iterate over all values of i
    for i in range(0, N):
 
        # If the index i is not visited already
        if (not visited[i]):
            countJumps(arr, N, i)
 
    # Print Answer
    for i in range(0, N):
        print(cntJumps[i], end=" ")
 
# Driver Code
if __name__ == "__main__":
 
    arr = [3, 12, 2, 7, 4, 10, 35, 5, 9, 27]
    N = len(arr)
    printCountJumps(arr, N)
 
    # This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
 
public class GFG {
 
// Stores the number of jumps
static int []cntJumps = new int[100];
 
// Stores if the current index is visited
static int []visited = new int[100];
 
// Recursive function to find the number of
// jumps to escape the array from index i
static int countJumps(int []arr, int N, int i)
{
   
    // If the answer for the current index is
    // already calculated
    if (cntJumps[i] != 0) {
        return cntJumps[i];
    }
 
    // If the current index is already visited
    if (visited[i] != 0) {
        return -1;
    }
 
    // Mark current index as visited
    visited[i] = 1;
 
    // If the array is escaped after the jump
    // from the current index
    if (arr[i] >= N) {
        cntJumps[i] = 1;
        return cntJumps[i];
    }
 
    // Recursive call for the next jump
    int val = countJumps(arr, N, arr[i]);
 
    // If it is impossible to escape the array
    if (val == -1)
        cntJumps[i] = -1;
    else
        cntJumps[i] = 1 + val;
 
    // Return answer
    return cntJumps[i];
}
 
// Function to print the number of jumps
// required to escape the array from
// ith index for all values of i in [0, N)
static void printCountJumps(int []arr, int N)
{
   
    // Initialize visited array as 0
    for (int i = 0; i < visited.Length; i++)
        visited[i] = 0;
         
    // Initialize cntJump array by 0
    for (int i = 0; i < cntJumps.Length; i++)
        cntJumps[i] = 0;
 
    // Loop to iterate over all values of i
    for (int i = 0; i < N; i++) {
 
        // If the index i is not visited already
        if (visited[i] == 0) {
            countJumps(arr, N, i);
        }
    }
 
    // Print Answer
    for (int i = 0; i < N; i++) {
        Console.Write(cntJumps[i] + " ");
    }
}
 
    // Driver Code
    public static void Main (string[] args)
    {
        int []arr = { 3, 12, 2, 7, 4, 10, 35, 5, 9, 27 };
        int N = arr.Length;
     
        printCountJumps(arr, N);
     
    }
}
 
// This code is contributed by AnkThon

Javascript

输出：

4 1 -1 3 -1 1 1 2 2 1

时间复杂度： O(N)
辅助空间： O(N)