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：
Input: arr[] = {1, 1, 1, 1, 1}
Output:
0 ⇾ 1 ⇾ 2 ⇾ 3 ⇾4
Explanation:
In every step, only one jump is allowed.
Therefore, only one possible path exists to reach end of the array.
Input: arr[] = {3, 3, 0, 2, 1, 2, 4, 2, 0, 0}
Output:
0 ⇾ 3 ⇾ 5 ⇾ 6 ⇾ 9
0 ⇾ 3 ⇾ 5 ⇾ 7 ⇾ 9
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 – 2到1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 akhilsainiC#
// 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