📌  相关文章
📜  需要最少跳转次数才能到达数组末尾的路径

📅  最后修改于: 2021-09-07 03:25:01             🧑  作者: Mango

给定一个数组arr[],其中每个元素表示从该元素向前可以进行的最大步数,任务是打印所有可能的路径,这些路径需要最少的跳转次数才能到达给定数组的末尾,从第一个数组元素。

注意:如果元素为 0,则不允许从该元素移动。

例子:

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:

  • 初始化大小为N的数组dp[] ,其中dp[i]存储从索引i到达数组arr[N – 1]末尾所需的最小跳转次数
  • 通过迭代索引N – 21来计算每个索引到达数组末尾所需的最小步数。对于每个索引,尝试可以从该索引中采取的所有可能步骤,即[1, arr[i]]
  • 在通过迭代[1, arr[i]]尝试所有可能的步骤时,对于每个索引,更新并存储dp[i + j]的最小值。
  • 初始化一个 Pair 类实例的队列,该队列存储当前位置的索引以及到目前为止到达该索引的路径。
  • 不断更新所需的最少步数,最后打印与所需步数对应的路径。

下面是上述方法的实现:

C++
// C++ program to implement the
// above approach
#include 
using namespace std;
 
// Pair Struct
struct Pair
{
     
    // Stores the current index
    int idx;
 
    // Stores the path
    // travelled so far
    string psf;
 
    // Stores the minimum jumps
    // required to reach the
    // last index from current index
    int jmps;
};
 
// Minimum jumps required to reach
// end of the array
void minJumps(int arr[], int dp[], int n)
{
    for(int i = 0; i < n; i++)
        dp[i] = INT_MAX;
 
    dp[n - 1] = 0;
 
    for(int i = n - 2; i >= 0; i--)
    {
         
        // Stores the maximum number
        // of steps that can be taken
        // from the current index
        int steps = arr[i];
        int min = INT_MAX;
 
        for(int j = 1;
                j <= steps && i + j < n;
                j++)
        {
             
            // Checking if index stays
            // within bounds
            if (dp[i + j] != INT_MAX &&
                dp[i + j] < min)
            {
                 
                // Stores the minimum
                // number of jumps
                // required to jump
                // from (i + j)-th index
                min = dp[i + j];
            }
        }
         
        if (min != INT_MAX)
            dp[i] = min + 1;
    }
}
 
// Function to find all possible
// paths to reach end of array
// requiring minimum number of steps
void possiblePath(int arr[], int dp[], int n)
{
    queue Queue;
    Pair p1 = { 0, "0", dp[0] };
    Queue.push(p1);
 
    while (Queue.size() > 0)
    {
        Pair tmp = Queue.front();
        Queue.pop();
 
        if (tmp.jmps == 0)
        {
            cout << tmp.psf << "\n";
            continue;
        }
 
        for(int step = 1;
                step <= arr[tmp.idx];
                step++)
        {
            if (tmp.idx + step < n &&
                tmp.jmps - 1 == dp[tmp.idx + step])
            {
                 
                // Storing the neighbours
                // of current index element
                string s1 = tmp.psf + " -> " +
                 to_string((tmp.idx + step));
                  
                Pair p2 = { tmp.idx + step, s1,
                           tmp.jmps - 1 };
                           
                Queue.push(p2);
            }
        }
    }
}
 
// Function to find the minimum steps
// and corresponding paths to reach
// end of an array
void Solution(int arr[], int dp[], int size)
{
     
    // dp[] array stores the minimum jumps
    // from each position to last position
    minJumps(arr, dp, size);
 
    possiblePath(arr, dp, size);
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 3, 0, 2, 1,
                  2, 4, 2, 0, 0 };
 
    int size = sizeof(arr) / sizeof(arr[0]);
    int dp[size];
 
    Solution(arr, dp, size);
}
 
// This code is contributed by akhilsaini


Java
// Java Program to implement the
// above approach
 
import java.util.*;
class GFG {
 
    // Pair Class instance
    public static class Pair {
        // Stores the current index
        int idx;
 
        // Stores the path
        // travelled so far
        String psf;
 
        // Stores the minimum jumps
        // required to reach the
        // last index from current index
        int jmps;
 
        // Constructor
        Pair(int idx, String psf, int jmps)
        {
            this.idx = idx;
            this.psf = psf;
            this.jmps = jmps;
        }
    }
 
    // Minimum jumps required to reach
    // end of the array
    public static int[] minJumps(int[] arr)
    {
        int dp[] = new int[arr.length];
 
        Arrays.fill(dp, Integer.MAX_VALUE);
 
        int n = dp.length;
 
        dp[n - 1] = 0;
 
        for (int i = n - 2; i >= 0; i--) {
            // Stores the maximum number
            // of steps that can be taken
            // from the current index
            int steps = arr[i];
            int min = Integer.MAX_VALUE;
 
            for (int j = 1; j <= steps && i + j < n; j++) {
                // Checking if index stays
                // within bounds
                if (dp[i + j] != Integer.MAX_VALUE
                    && dp[i + j] < min) {
                    // Stores the minimum
                    // number of jumps
                    // required to jump
                    // from (i + j)-th index
                    min = dp[i + j];
                }
            }
 
            if (min != Integer.MAX_VALUE)
                dp[i] = min + 1;
        }
        return dp;
    }
 
    // Function to find all possible
    // paths to reach end of array
    // requiring minimum number of steps
    public static void possiblePath(
        int[] arr, int[] dp)
    {
 
        Queue queue = new LinkedList<>();
        queue.add(new Pair(0, "" + 0, dp[0]));
 
        while (queue.size() > 0) {
            Pair tmp = queue.remove();
 
            if (tmp.jmps == 0) {
                System.out.println(tmp.psf);
                continue;
            }
 
            for (int step = 1;
                 step <= arr[tmp.idx];
                 step++) {
 
                if (tmp.idx + step < arr.length
                    && tmp.jmps - 1 == dp[tmp.idx + step]) {
                    // Storing the neighbours
                    // of current index element
                    queue.add(new Pair(
                        tmp.idx + step,
                        tmp.psf + " -> " + (tmp.idx + step),
                        tmp.jmps - 1));
                }
            }
        }
    }
 
    // Function to find the minimum steps
    // and corresponding paths to reach
    // end of an array
    public static void Solution(int arr[])
    {
        // Stores the minimum jumps from
        // each position to last position
        int dp[] = minJumps(arr);
 
        possiblePath(arr, dp);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = { 3, 3, 0, 2, 1,
                      2, 4, 2, 0, 0 };
        int size = arr.length;
        Solution(arr);
    }
}


Python3
# Python3 program to implement the
# above approach
from queue import Queue
import sys
 
# Pair Class instance
class Pair(object):
     
    # Stores the current index
    idx = 0
 
    # Stores the path
    # travelled so far
    psf = ""
 
    # Stores the minimum jumps
    # required to reach the
    # last index from current index
    jmps = 0
 
    # Constructor
    def __init__(self, idx, psf, jmps):
         
        self.idx = idx
        self.psf = psf
        self.jmps = jmps
 
# Minimum jumps required to reach
# end of the array
def minJumps(arr):
 
    MAX_VALUE = sys.maxsize
    dp = [MAX_VALUE for i in range(len(arr))]
 
    n = len(dp)
 
    dp[n - 1] = 0
 
    for i in range(n - 2, -1, -1):
         
        # Stores the maximum number
        # of steps that can be taken
        # from the current index
        steps = arr[i]
        minimum = MAX_VALUE
 
        for j in range(1, steps + 1, 1):
            if i + j >= n:
                break
             
            # Checking if index stays
            # within bounds
            if ((dp[i + j] != MAX_VALUE) and
                (dp[i + j] < minimum)):
                     
                # Stores the minimum
                # number of jumps
                # required to jump
                # from (i + j)-th index
                minimum = dp[i + j]
 
        if minimum != MAX_VALUE:
            dp[i] = minimum + 1
             
    return dp
 
# Function to find all possible
# paths to reach end of array
# requiring minimum number of steps
def possiblePath(arr, dp):
 
    queue = Queue(maxsize = 0)
    p1 = Pair(0, "0", dp[0])
    queue.put(p1)
 
    while queue.qsize() > 0:
        tmp = queue.get()
 
        if tmp.jmps == 0:
            print(tmp.psf)
            continue
 
        for step in range(1, arr[tmp.idx] + 1, 1):
            if ((tmp.idx + step < len(arr)) and
               (tmp.jmps - 1 == dp[tmp.idx + step])):
               
                # Storing the neighbours
                # of current index element
                p2 = Pair(tmp.idx + step, tmp.psf +
                           " -> " + str((tmp.idx + step)),
                         tmp.jmps - 1)
                          
                queue.put(p2)
 
# Function to find the minimum steps
# and corresponding paths to reach
# end of an array
def Solution(arr):
     
    # Stores the minimum jumps from
    # each position to last position
    dp = minJumps(arr)
     
    possiblePath(arr, dp)
 
# Driver Code
if __name__ == "__main__":
     
    arr = [ 3, 3, 0, 2, 1,
            2, 4, 2, 0, 0 ]
    size = len(arr)
 
    Solution(arr)
 
# This code is contributed by akhilsaini


C#
// C# program to implement the
// above approach
using System;
using System.Collections;
 
// Pair Struct
public struct Pair
{
     
    // Stores the current index
    public int idx;
 
    // Stores the path
    // travelled so far
    public string psf;
 
    // Stores the minimum jumps
    // required to reach the
    // last index from current index
    public int jmps;
 
    // Constructor
    public Pair(int idx, String psf, int jmps)
    {
        this.idx = idx;
        this.psf = psf;
        this.jmps = jmps;
    }
}
 
class GFG{
 
// Minimum jumps required to reach
// end of the array
public static int[] minJumps(int[] arr)
{
    int[] dp = new int[arr.Length];
    int n = dp.Length;
 
    for(int i = 0; i < n; i++)
        dp[i] = int.MaxValue;
 
    dp[n - 1] = 0;
 
    for(int i = n - 2; i >= 0; i--)
    {
         
        // Stores the maximum number
        // of steps that can be taken
        // from the current index
        int steps = arr[i];
        int min = int.MaxValue;
 
        for(int j = 1;
                j <= steps && i + j < n;
                j++)
        {
             
            // Checking if index stays
            // within bounds
            if (dp[i + j] != int.MaxValue &&
                dp[i + j] < min)
            {
                 
                // Stores the minimum
                // number of jumps
                // required to jump
                // from (i + j)-th index
                min = dp[i + j];
            }
        }
 
        if (min != int.MaxValue)
            dp[i] = min + 1;
    }
    return dp;
}
 
// Function to find all possible
// paths to reach end of array
// requiring minimum number of steps
public static void possiblePath(int[] arr,
                                int[] dp)
{
    Queue queue = new Queue();
    queue.Enqueue(new Pair(0, "0", dp[0]));
 
    while (queue.Count > 0)
    {
        Pair tmp = (Pair)queue.Dequeue();
 
        if (tmp.jmps == 0)
        {
            Console.WriteLine(tmp.psf);
            continue;
        }
 
        for(int step = 1;
                step <= arr[tmp.idx];
                step++)
        {
            if (tmp.idx + step < arr.Length &&
               tmp.jmps - 1 == dp[tmp.idx + step])
            {
                 
                // Storing the neighbours
                // of current index element
 
                queue.Enqueue(new Pair(
                    tmp.idx + step,
                    tmp.psf + " -> " +
                   (tmp.idx + step),
                   tmp.jmps - 1));
            }
        }
    }
}
 
// Function to find the minimum steps
// and corresponding paths to reach
// end of an array
public static void Solution(int[] arr)
{
     
    // Stores the minimum jumps from
    // each position to last position
    int[] dp = minJumps(arr);
 
    possiblePath(arr, dp);
}
 
// Driver Code
static public void Main()
{
    int[] arr = { 3, 3, 0, 2, 1,
                  2, 4, 2, 0, 0 };
    int size = arr.Length;
     
    Solution(arr);
}
}
 
// This code is contributed by akhilsaini


输出:
0 -> 3 -> 5 -> 6 -> 9
0 -> 3 -> 5 -> 7 -> 9








时间复杂度: O(N 2 )
辅助空间: O(N)

如果您想与行业专家一起参加直播课程,请参阅Geeks Classes Live