最低成本路径| DP-6

给定成本矩阵cost [] []和cost [] []中的位置(m，n)，编写一个函数，返回从(0，0)到(m，n)的最小成本路径的成本。矩阵的每个单元代表遍历该单元的成本。到达(m，n)的路径的总成本是该路径上所有成本(包括源和目标)的总和。您只能遍历给定单元格(即给定单元格(i，j)，单元格(i + 1，j)，(i，j + 1)和(i + 1)的向下，向右和对角线较低的单元格，j + 1)可以遍历。您可以假设所有成本都是正整数。

例如，在下图中，到(2，2)的最小成本路径是什么?

下图突出显示了成本最低的路径。路径是(0，0)–>(0，1)–>(1，2)–>(2，2)。路径成本为8(1 + 2 + 2 + 3)。

1)最佳子结构
到达(m，n)的路径必须通过以下三个单元之一：(m-1，n-1)或(m-1，n)或(m，n-1)。因此，达到(m，n)的最小成本可以写为“ 3个单元中的最小值加上cost [m] [n]”。
minCost(m，n)=分钟(minCost(m-1，n-1)，minCost(m-1，n)，minCost(m，n-1))+ cost [m] [n]

2)重叠子问题
以下是MCP(最小成本路径)问题的简单递归实现。该实现仅遵循上述递归结构。

C++

// A Naive recursive implementation
// of MCP(Minimum Cost Path) problem
#include 
using namespace std;
 
#define R 3
#define C 3
 
int min(int x, int y, int z);
 
// Returns cost of minimum cost path
// from (0,0) to (m, n) in mat[R][C]
int minCost(int cost[R][C], int m, int n)
{
    if (n < 0 || m < 0)
        return INT_MAX;
    else if (m == 0 && n == 0)
        return cost[m][n];
    else
        return cost[m][n] +
         min(minCost(cost, m - 1, n - 1),
             minCost(cost, m - 1, n),
             minCost(cost, m, n - 1));
}
 
// A utility function that returns
// minimum of 3 integers
int min(int x, int y, int z)
{
    if (x < y)
        return (x < z) ? x : z;
    else
        return (y < z) ? y : z;
}
 
// Driver code
int main()
{
    int cost[R][C] = { { 1, 2, 3 },
                       { 4, 8, 2 },
                       { 1, 5, 3 } };
                        
    cout << minCost(cost, 2, 2) << endl;
     
    return 0;
}
 
// This code is contributed by nikhilchhipa9

C

/* A Naive recursive implementation of MCP(Minimum Cost Path) problem */
#include
#include
#define R 3
#define C 3
 
int min(int x, int y, int z);
 
/* Returns cost of minimum cost path from (0,0) to (m, n) in mat[R][C]*/
int minCost(int cost[R][C], int m, int n)
{
   if (n < 0 || m < 0)
      return INT_MAX;
   else if (m == 0 && n == 0)
      return cost[m][n];
   else
      return cost[m][n] + min( minCost(cost, m-1, n-1),
                               minCost(cost, m-1, n),
                               minCost(cost, m, n-1) );
}
 
/* A utility function that returns minimum of 3 integers */
int min(int x, int y, int z)
{
   if (x < y)
      return (x < z)? x : z;
   else
      return (y < z)? y : z;
}
 
/* Driver program to test above functions */
int main()
{
   int cost[R][C] = { {1, 2, 3},
                      {4, 8, 2},
                      {1, 5, 3} };
   printf(" %d ", minCost(cost, 2, 2));
   return 0;
}

Java

/* A Naive recursive implementation of
MCP(Minimum Cost Path) problem */
public class GFG {
 
    /* A utility function that returns
    minimum of 3 integers */
    static int min(int x, int y, int z)
    {
        if (x < y)
            return (x < z) ? x : z;
        else
            return (y < z) ? y : z;
    }
     
    /* Returns cost of minimum cost path
    from (0,0) to (m, n) in mat[R][C]*/
    static int minCost(int cost[][], int m,
                                     int n)
    {
        if (n < 0 || m < 0)
            return Integer.MAX_VALUE;
        else if (m == 0 && n == 0)
            return cost[m][n];
        else
            return cost[m][n] +
                min( minCost(cost, m-1, n-1),
                     minCost(cost, m-1, n),
                     minCost(cost, m, n-1) );
    }
 
    // Driver code
    public static void main(String args[])
    {
         
        int cost[][] = { {1, 2, 3},
                         {4, 8, 2},
                         {1, 5, 3} };
                          
        System.out.print(minCost(cost, 2, 2));
    }
}
 
// This code is contributed by Sam007

Python3

# A Naive recursive implementation of MCP(Minimum Cost Path) problem
R = 3
C = 3
import sys
 
# Returns cost of minimum cost path from (0,0) to (m, n) in mat[R][C]
def minCost(cost, m, n):
    if (n < 0 or m < 0):
        return sys.maxsize
    elif (m == 0 and n == 0):
        return cost[m][n]
    else:
        return cost[m][n] + min( minCost(cost, m-1, n-1),
                                minCost(cost, m-1, n),
                                minCost(cost, m, n-1) )
 
#A utility function that returns minimum of 3 integers */
def min(x, y, z):
    if (x < y):
        return x if (x < z) else z
    else:
        return y if (y < z) else z
 
 
# Driver program to test above functions
cost= [ [1, 2, 3],
        [4, 8, 2],
        [1, 5, 3] ]
print(minCost(cost, 2, 2))
 
# This code is contributed by
# Smitha Dinesh Semwal

C#

/* A Naive recursive implementation of
MCP(Minimum Cost Path) problem */
using System;
 
class GFG
{
 
    /* A utility function that
    returns minimum of 3 integers */
    static int min(int x,
                   int y, int z)
    {
        if (x < y)
            return ((x < z) ? x : z);
        else
            return ((y < z) ? y : z);
    }
     
    /* Returns cost of minimum
    cost path from (0,0) to
    (m, n) in mat[R][C]*/
    static int minCost(int [,]cost,
                       int m , int n)
    {
        if (n < 0 || m < 0)
            return int.MaxValue;
        else if (m == 0 && n == 0)
            return cost[m, n];
        else
            return cost[m, n] +
                   min(minCost(cost, m - 1, n - 1),
                   minCost(cost, m - 1, n),
                   minCost(cost, m, n - 1) );
    }
 
    // Driver code
    public static void Main()
    {
         
        int [,]cost = {{1, 2, 3},
                       {4, 8, 2},
                       {1, 5, 3}};
                         
        Console.Write(minCost(cost, 2, 2));
    }
}
 
// This code is contributed
// by shiv_bhakt.

Javascript

 
// A Naive recursive implementation of
// MCP(Minimum Cost Path) problem
 
// A utility function that returns
// minimum of 3 integers
function min(x, y, z)
{
    if (x < y)
        return (x < z) ? x : z;
    else
        return (y < z) ? y : z;
}
 
// Returns cost of minimum cost path
// from (0,0) to (m, n) in mat[R][C]
function minCost(cost, m, n)
{
    if (n < 0 || m < 0)
        return Number.MAX_VALUE;
    else if (m == 0 && n == 0)
        return cost[m][n];
    else
        return cost[m][n] + min(minCost(cost, m - 1, n - 1),
                                minCost(cost, m - 1, n),
                                minCost(cost, m, n - 1));
}
 
// Driver code
var cost = [ [ 1, 2, 3 ],
             [ 4, 8, 2 ],
             [ 1, 5, 3 ] ];
 
document.write(minCost(cost, 2, 2));
 
// This code is contributed by gauravrajput1
 

---

输出

8

应该注意的是，上述函数一次又一次地计算相同的子问题。参见下面的递归树，有很多节点出现不止一次。这种幼稚的递归解决方案的时间复杂度是指数级的，并且非常慢。

mC refers to minCost()
                                    mC(2, 2)
                          /            |           \
                         /             |            \             
                 mC(1, 1)           mC(1, 2)             mC(2, 1)
              /     |     \       /     |     \           /     |     \ 
             /      |      \     /      |      \         /      |       \
       mC(0,0) mC(0,1) mC(1,0) mC(0,1) mC(0,2) mC(1,1) mC(1,0) mC(1,1) mC(2,0)

因此，MCP问题具有动态编程问题的两个属性(请参阅此内容)。像其他典型的动态编程(DP)问题一样，可以通过以自下而上的方式构造临时数组tc [] []来避免相同子问题的重新计算。

C++

/* Dynamic Programming implementation of MCP problem */
#include 
#include
#define R 3
#define C 3
using namespace std;
int min(int x, int y, int z);
 
int minCost(int cost[R][C], int m, int n)
{
     int i, j;
 
     // Instead of following line, we can use int tc[m+1][n+1] or
     // dynamically allocate memory to save space. The following line is
     // used to keep the program simple and make it working on all compilers.
     int tc[R][C]; 
 
     tc[0][0] = cost[0][0];
 
     /* Initialize first column of total cost(tc) array */
     for (i = 1; i <= m; i++)
        tc[i][0] = tc[i - 1][0] + cost[i][0];
 
     /* Initialize first row of tc array */
     for (j = 1; j <= n; j++)
        tc[0][j] = tc[0][j - 1] + cost[0][j];
 
     /* Construct rest of the tc array */
     for (i = 1; i <= m; i++)
        for (j = 1; j <= n; j++)
            tc[i][j] = min(tc[i - 1][j - 1],
                           tc[i - 1][j],
                           tc[i][j - 1]) + cost[i][j];
 
     return tc[m][n];
}
 
/* A utility function that returns minimum of 3 integers */
int min(int x, int y, int z)
{
   if (x < y)
      return (x < z)? x : z;
   else
      return (y < z)? y : z;
}
 
/* Driver code*/
int main()
{
   int cost[R][C] = { {1, 2, 3},
                      {4, 8, 2},
                      {1, 5, 3} };
   cout << "  " << minCost(cost, 2, 2);
   return 0;
}
 
// This code is contributed by shivanisinghss2110

C

/* Dynamic Programming implementation of MCP problem */
#include
#include
#define R 3
#define C 3
 
int min(int x, int y, int z);
 
int minCost(int cost[R][C], int m, int n)
{
     int i, j;
 
     // Instead of following line, we can use int tc[m+1][n+1] or
     // dynamically allocate memory to save space. The following line is
     // used to keep the program simple and make it working on all compilers.
     int tc[R][C]; 
 
     tc[0][0] = cost[0][0];
 
     /* Initialize first column of total cost(tc) array */
     for (i = 1; i <= m; i++)
        tc[i][0] = tc[i-1][0] + cost[i][0];
 
     /* Initialize first row of tc array */
     for (j = 1; j <= n; j++)
        tc[0][j] = tc[0][j-1] + cost[0][j];
 
     /* Construct rest of the tc array */
     for (i = 1; i <= m; i++)
        for (j = 1; j <= n; j++)
            tc[i][j] = min(tc[i-1][j-1],
                           tc[i-1][j],
                           tc[i][j-1]) + cost[i][j];
 
     return tc[m][n];
}
 
/* A utility function that returns minimum of 3 integers */
int min(int x, int y, int z)
{
   if (x < y)
      return (x < z)? x : z;
   else
      return (y < z)? y : z;
}
 
/* Driver program to test above functions */
int main()
{
   int cost[R][C] = { {1, 2, 3},
                      {4, 8, 2},
                      {1, 5, 3} };
   printf(" %d ", minCost(cost, 2, 2));
   return 0;
}

Java

/* Java program for Dynamic Programming implementation
   of Min Cost Path problem */
import java.util.*;
 
class MinimumCostPath
{
    /* A utility function that returns minimum of 3 integers */
    private static int min(int x, int y, int z)
    {
        if (x < y)
            return (x < z)? x : z;
        else
            return (y < z)? y : z;
    }
 
    private static int minCost(int cost[][], int m, int n)
    {
        int i, j;
        int tc[][]=new int[m+1][n+1];
 
        tc[0][0] = cost[0][0];
 
        /* Initialize first column of total cost(tc) array */
        for (i = 1; i <= m; i++)
            tc[i][0] = tc[i-1][0] + cost[i][0];
 
        /* Initialize first row of tc array */
        for (j = 1; j <= n; j++)
            tc[0][j] = tc[0][j-1] + cost[0][j];
 
        /* Construct rest of the tc array */
        for (i = 1; i <= m; i++)
            for (j = 1; j <= n; j++)
                tc[i][j] = min(tc[i-1][j-1],
                               tc[i-1][j],
                               tc[i][j-1]) + cost[i][j];
 
        return tc[m][n];
    }
 
    /* Driver program to test above functions */
    public static void main(String args[])
    {
        int cost[][]= {{1, 2, 3},
                       {4, 8, 2},
                       {1, 5, 3}};
        System.out.println(minCost(cost,2,2));
    }
}
// This code is contributed by Pankaj Kumar

Python

# Dynamic Programming Python implementation of Min Cost Path
# problem
R = 3
C = 3
 
def minCost(cost, m, n):
 
    # Instead of following line, we can use int tc[m+1][n+1] or
    # dynamically allocate memoery to save space. The following
    # line is used to keep te program simple and make it working
    # on all compilers.
    tc = [[0 for x in range(C)] for x in range(R)]
 
    tc[0][0] = cost[0][0]
 
    # Initialize first column of total cost(tc) array
    for i in range(1, m+1):
        tc[i][0] = tc[i-1][0] + cost[i][0]
 
    # Initialize first row of tc array
    for j in range(1, n+1):
        tc[0][j] = tc[0][j-1] + cost[0][j]
 
    # Construct rest of the tc array
    for i in range(1, m+1):
        for j in range(1, n+1):
            tc[i][j] = min(tc[i-1][j-1], tc[i-1][j], tc[i][j-1]) + cost[i][j]
 
    return tc[m][n]
 
# Driver program to test above functions
cost = [[1, 2, 3],
        [4, 8, 2],
        [1, 5, 3]]
print(minCost(cost, 2, 2))
 
# This code is contributed by Bhavya Jain

C＃

// C# program for Dynamic Programming implementation
// of Min Cost Path problem
using System;
 
class GFG
{
    // A utility function that
    // returns minimum of 3 integers
    private static int min(int x, int y, int z)
    {
        if (x < y)
            return (x < z)? x : z;
        else
            return (y < z)? y : z;
    }
 
    private static int minCost(int [,]cost, int m, int n)
    {
        int i, j;
        int [,]tc=new int[m+1,n+1];
 
        tc[0,0] = cost[0,0];
 
        /* Initialize first column of total cost(tc) array */
        for (i = 1; i <= m; i++)
            tc[i, 0] = tc[i - 1, 0] + cost[i, 0];
 
        /* Initialize first row of tc array */
        for (j = 1; j <= n; j++)
            tc[0, j] = tc[0, j - 1] + cost[0, j];
 
        /* Construct rest of the tc array */
        for (i = 1; i <= m; i++)
            for (j = 1; j <= n; j++)
                tc[i, j] = min(tc[i - 1, j - 1],
                            tc[i - 1, j],
                            tc[i, j - 1]) + cost[i, j];
 
        return tc[m, n];
    }
 
    // Driver program
    public static void Main()
    {
        int [,]cost= {{1, 2, 3},
                    {4, 8, 2},
                    {1, 5, 3}};
        Console.Write(minCost(cost,2,2));
    }
}
 
// This code is contributed by Sam007.

Java脚本

    // Javascript program for Dynamic Programming implementation
    // of Min Cost Path problem
     
    /* A utility function that returns minimum of 3 integers */
    function min(x, y, z)
    {
        if (x < y)
            return (x < z)? x : z;
        else
            return (y < z)? y : z;
    }
  
    function minCost(cost, m, n)
    {
        let i, j;
        let tc = new Array(m+1);
         
        for(let k = 0; k < m + 1; k++)
        {
            tc[k] = new Array(n+1);
        }
  
        tc[0][0] = cost[0][0];
  
        /* Initialize first column of total cost(tc) array */
        for (i = 1; i <= m; i++)
            tc[i][0] = tc[i-1][0] + cost[i][0];
  
        /* Initialize first row of tc array */
        for (j = 1; j <= n; j++)
            tc[0][j] = tc[0][j-1] + cost[0][j];
  
        /* Construct rest of the tc array */
        for (i = 1; i <= m; i++)
            for (j = 1; j <= n; j++)
                tc[i][j] = Math.min(tc[i-1][j-1],
                               tc[i-1][j],
                               tc[i][j-1]) + cost[i][j];
  
        return tc[m][n];
    }
       
    let cost = [[1, 2, 3],
                [4, 8, 2],
                 [1, 5, 3]];
    document.write(minCost(cost,2,2));
         

输出

8

DP实现的时间复杂度为O(mn)，这比朴素的递归实现要好得多。

空间优化：想法是使用相同的给定数组存储子问题的解决方案。

C++

#include 
#define endl "\n"
using namespace std;
 
const int row = 3;
const int col = 3;
 
int minCost(int cost[row][col]) {
 
    // for 1st column
    for (int i=1 ; i

Java

// Java program for the
// above approach
import java.util.*;
class GFG{
 
static int row = 3;
static int col = 3;
 
static int minCost(int cost[][])
{
  // for 1st column
  for (int i = 1; i < row; i++)
  {
    cost[i][0] += cost[i - 1][0];
  }
 
  // for 1st row
  for (int j = 1; j < col; j++)
  {
    cost[0][j] += cost[0][j - 1];
  }
 
  // for rest of the 2d matrix
  for (int i = 1; i < row; i++)
  {
    for (int j = 1; j < col; j++)
    {
      cost[i][j] += Math.min(cost[i - 1][j - 1],
                    Math.min(cost[i - 1][j],
                             cost[i][j - 1]));
    }
  }
 
  // Returning the value in
  // last cell
  return cost[row - 1][col - 1];
}
   
// Driver code 
public static void  main(String[] args)
{    
  int cost[][] = {{1, 2, 3},
                  {4, 8, 2},
                  {1, 5, 3} };
  System.out.print(minCost(cost) + "\n");
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the
# above approach
 
def minCost(cost, row, col):
     
    # For 1st column
    for i in range(1, row):
        cost[i][0] += cost[i - 1][0]
 
    # For 1st row
    for j in range(1, col):
        cost[0][j] += cost[0][j - 1]
 
    # For rest of the 2d matrix
    for i in range(1, row):
        for j in range(1, col):
            cost[i][j] += (min(cost[i - 1][j - 1],
                           min(cost[i - 1][j],
                               cost[i][j - 1])))
 
    # Returning the value in
    # last cell
    return cost[row - 1][col - 1]
 
# Driver code
if __name__ == '__main__':
     
    row = 3
    col = 3
     
    cost = [ [ 1, 2, 3 ],
             [ 4, 8, 2 ],
             [ 1, 5, 3 ] ]
              
    print(minCost(cost, row, col));
 
# This code is contributed by Amit Katiyar

C＃

// C# program for the
// above approach
using System;
class GFG{
 
static int row = 3;
static int col = 3;
 
static int minCost(int [,]cost)
{
  // for 1st column
  for (int i = 1; i < row; i++)
  {
    cost[i, 0] += cost[i - 1, 0];
  }
 
  // for 1st row
  for (int j = 1; j < col; j++)
  {
    cost[0, j] += cost[0, j - 1];
  }
 
  // for rest of the 2d matrix
  for (int i = 1; i < row; i++)
  {
    for (int j = 1; j < col; j++)
    {
      cost[i,j] += Math.Min(cost[i - 1,
                                 j - 1],
                   Math.Min(cost[i - 1, j],
                            cost[i, j - 1]));
    }
  }
 
  // Returning the value in
  // last cell
  return cost[row - 1, col - 1];
}
   
// Driver code 
public static void  Main(String[] args)
{    
  int [,]cost = {{1, 2, 3},
                 {4, 8, 2},
                 {1, 5, 3} };
  Console.Write(minCost(cost) + "\n");
}
}
 
// This code is contributed by Rajput-Ji

Java脚本

 
// javascript program for the
// above approach
 
var row = 3;
var col = 3;
 
function minCost(cost)
{
  // for 1st column
  for (i = 1; i < row; i++)
  {
    cost[i][0] += cost[i - 1][0];
  }
 
  // for 1st row
  for (j = 1; j < col; j++)
  {
    cost[0][j] += cost[0][j - 1];
  }
 
  // for rest of the 2d matrix
  for (i = 1; i < row; i++)
  {
    for (j = 1; j < col; j++)
    {
      cost[i][j] += Math.min(cost[i - 1][j - 1],
                    Math.min(cost[i - 1][j],
                             cost[i][j - 1]));
    }
  }
 
  // Returning the value in
  // last cell
  return cost[row - 1][col - 1];
}
   
// Driver code 
var cost = [[1, 2, 3],
[4, 8, 2],
[1, 5, 3] ];
document.write(minCost(cost) + '<br>');
 
// This code is contributed by 29AjayKumar
 

输出

8

替代解决方案
我们还可以使用Dijkstra的最短路径算法。下面是该方法的实现：

C++

/* Minimum Cost Path using Dijkstra’s shortest path
   algorithm with Min Heap by dinglizeng */
#include 
#include 
#include 
using namespace std;
 
/* define the number of rows and the number of columns */
#define R 4
#define C 5
 
/* 8 possible moves */
int dx[] = {1,-1, 0, 0, 1, 1,-1,-1};
int dy[] = {0, 0, 1,-1, 1,-1, 1,-1};
 
/* The data structure to store the coordinates of \\
  the unit square and the cost of path from the top left. */
struct Cell{
    int x;
    int y;
    int cost;
};
 
/* The compare class to be used by a Min Heap.
 * The greater than condition is used as this
   is for a Min Heap based on priority_queue.
 */
class mycomparison
{
public:
  bool operator() (const Cell &lhs, const Cell &rhs) const
  {
    return (lhs.cost > rhs.cost);
  }
};
 
/* To verify whether a move is within the boundary. */
bool isSafe(int x, int y){
    return x >= 0 && x < R && y >= 0 && y < C;
}
 
/* This solution is based on Dijkstra’s shortest path algorithm
 * For each unit square being visited, we examine all
    possible next moves in 8 directions,
 *    calculate the accumulated cost of path for each
     next move, adjust the cost of path of the adjacent
     units to the minimum as needed.
 *    then add the valid next moves into a Min Heap.
 * The Min Heap pops out the next move with the minimum
   accumulated cost of path.
 * Once the iteration reaches the last unit at the lower
   right corner, the minimum cost path will be returned.
 */
int minCost(int cost[R][C], int m, int n) {
 
    /* the array to store the accumulated cost
       of path from top left corner */
    int dp[R][C];
 
    /* the array to record whether a unit
       square has been visited */
    bool visited[R][C];
    
    /* Initialize these two arrays, set path cost
      to maximum integer value, each unit as not visited */
    for(int i = 0; i < R; i++) {
        for(int j = 0; j < C; j++) {
            dp[i][j] = INT_MAX;
            visited[i][j] = false;
        }
    }
     
    /* Define a reverse priority queue.
     * Priority queue is a heap based implementation.
     * The default behavior of a priority queue is
        to have the maximum element at the top.
     * The compare class is used in the definition of the Min Heap.
     */
    priority_queue, mycomparison> pq;
     
    /* initialize the starting top left unit with the
      cost and add it to the queue as the first move. */
    dp[0][0] = cost[0][0];
    pq.push({0, 0, cost[0][0]});
     
    while(!pq.empty()) {
 
        /* pop a move from the queue, ignore the units
           already visited */
        Cell cell = pq.top();
        pq.pop();
        int x = cell.x;
        int y = cell.y;
        if(visited[x][y]) continue;
 
        /* mark the current unit as visited */
        visited[x][y] = true;
 
        /* examine all non-visited adjacent units in 8 directions 
         * calculate the accumulated cost of path for
           each next move from this unit,
         * adjust the cost of path for each next adjacent
           units to the minimum if possible.
         */
        for(int i = 0; i < 8; i++) {
            int next_x = x + dx[i];
            int next_y = y + dy[i];
            if(isSafe(next_x, next_y) && !visited[next_x][next_y]) {
                dp[next_x][next_y] = min(dp[next_x][next_y],
                   dp[x][y] + cost[next_x][next_y]);
                pq.push({next_x, next_y, dp[next_x][next_y]});
            }
        }
    }
 
    /* return the minimum cost path at the lower
       right corner */
    return dp[m][n];       
}
 
/* Driver program to test above functions */
int main()
{
   int cost[R][C] = { {1, 8, 8, 1, 5},
                      {4, 1, 1, 8, 1},
                      {4, 2, 8, 8, 1},
                      {1, 5, 8, 8, 1} };
   printf(" %d ", minCost(cost, 3, 4));
   return 0;
}

输出

7

与以最小路径成本寻找节点的全扫描相比，在该解决方案中使用反向优先级队列可以降低时间复杂度。不考虑使用优先级队列的情况下，DP实现的总体时间复杂度为O(mn)，这比Naive Recursive实现要好得多。

?list = PLqM7alHXFySEQDk2MDfbwEdjd2svVJH9p

询问：亚马逊