如何使用堆栈而不是递归？

Question

我想使用堆栈而不是递归。我回顾了类似的答案，他们似乎很好，但我真的不明白。我所做的只是使代码变得更糟。 我想在以下代码中使用堆栈：http://www.geeksforgeeks.org/backtracking-set-3-n-queen-problem/
下面的代码通过递归实现n皇后问题递归，它工作正常。我只需要使用堆栈而不是递归。我认为回溯算法是使用堆栈创建递归的好主意。

任何帮助将不胜感激。

#define N 4 
#include<stdio.h> 

/* A utility function to print solution */ 
void printSolution(int board[N][N]) 
{ 
    for (int i = 0; i < N; i++) 
    { 
     for (int j = 0; j < N; j++) 
      printf(" %d ", board[i][j]); 
     printf("\n"); 
    } 
} 

/* A utility function to check if a queen can be placed on board[row][col] 
    Note that this function is called when "col" queens are already placeed 
    in columns from 0 to col -1. So we need to check only left side for 
    attacking queens */ 
bool isSafe(int board[N][N], int row, int col) 
{ 
    int i, j; 

    /* Check this row on left side */ 
    for (i = 0; i < col; i++) 
    { 
     if (board[row][i]) 
      return false; 
    } 

    /* Check upper diagonal on left side */ 
    for (i = row, j = col; i >= 0 && j >= 0; i--, j--) 
    { 
     if (board[i][j]) 
      return false; 
    } 

    /* Check lower diagonal on left side */ 
    for (i = row, j = col; j >= 0 && i < N; i++, j--) 
    { 
     if (board[i][j]) 
      return false; 
    } 

    return true; 
} 

/* A recursive utility function to solve N Queen problem */ 
bool solveNQUtil(int board[N][N], int col) 
{ 
    /* base case: If all queens are placed then return true */ 
    if (col >= N) 
     return true; 

    /* Consider this column and try placing this queen in all rows 
     one by one */ 
    for (int i = 0; i < N; i++) 
    { 
     /* Check if queen can be placed on board[i][col] */ 
     if (isSafe(board, i, col)) 
     { 
      /* Place this queen in board[i][col] */ 
      board[i][col] = 1; 

      /* recur to place rest of the queens */ 
      if (solveNQUtil(board, col + 1) == true) 
       return true; 

      /* If placing queen in board[i][col] doesn't lead to a solution 
       then remove queen from board[i][col] */ 
      board[i][col] = 0; // BACKTRACK 
     } 
    } 

    /* If queen can not be place in any row in this colum col 
     then return false */ 
    return false; 
} 

/* This function solves the N Queen problem using Backtracking. It mainly uses 
solveNQUtil() to solve the problem. It returns false if queens cannot be placed, 
otherwise return true and prints placement of queens in the form of 1s. Please 
note that there may be more than one solutions, this function prints one of the 
feasible solutions.*/ 
bool solveNQ() 
{ 
    int board[N][N] = { {0, 0, 0, 0}, 
     {0, 0, 0, 0}, 
     {0, 0, 0, 0}, 
     {0, 0, 0, 0} 
    }; 

    if (solveNQUtil(board, 0) == false) 
    { 
     printf("Solution does not exist"); 
     return false; 
    } 

    printSolution(board); 
    return true; 
} 

// driver program to test above function 
int main() 
{ 
    solveNQ(); 

    getchar(); 
    return 0; 
} 

Answer 1

考虑到在每次递归时都要测试下一列。你的电路板上已经有了一些栈，col是你的堆栈指针。

考虑到你只有一个女王，你甚至可以减小所需的头部空间的大小。

const int size=5; // your board size 
int isSafe(const int pos[size],const int col,const int row) { 
    for(int i=0;i<col;i++) { //check up to the stack pointer (anything beyond is garbage) 
     if(pos[i]==row){ //check on same row 
      return 0; 
     } 
     //check diagonals 
     int delta=col-i; 
     if(pos[i]==(row+delta)) { 
      return 0; 
     } 
     if(pos[i]==(row-delta)) { 
      return 0; 
     } 
    } 
    return 1; 
} 

int solveNQ() { 
    int pos[size]; 
    int col=0; 
    int row=0; 
    while(col<size) { 
     if(isSafe(pos,col,row)) { 
      pos[col]=row; 
      col++; 
      row=0; 
     }else{ 
      row++; 
      while(row==size) { // rollback to previous while end of col 
       if(col==0) { 
        // we seem to have found no reasonable solution 
        // as we rolled back further 
        return 0; 
       } 
       col--; 
       row=pos[col]+1; // pop previous position and try the next one 
      } 
     } 
    } 
    return 1; 
} 
int main() { 
    solveNQ(); 
    return 0; 
} 

这是在伪递归的方式