八皇后问题，递归和非递归解法

又一个经典问题，貌似很多公司笔试都考过这个问题的非递归程序求解。下面是我的递归与非递归程序，个人感觉程序逻辑还是比较清晰:)。

int board[8][8];
int cnt = 0;

bool isValid(int i, int j)
{
    int k;
    for(k = 0; k < 8; ++k)
        if(k != j && board[i][k]) return false;
    for(k = 0; k < 8; ++k)
        if(k != i && board[k][j]) return false;
    for(k = 1; k < 8 && i-k >= 0 && j-k >= 0; ++k)
        if(board[i-k][j-k]) return false;
    for(k = 1; k < 8 && i-k >= 0 && j+k < 8; ++k)
        if(board[i-k][j+k]) return false;
    return true;
}

void queen_rec(int row)
{
    if(row == 8)
    {
        for(int i = 0; i < 8; ++i)
            for(int j = 0; j < 8; ++j)
                if(board[i][j]) printf("%d %d\n", i, j);
        printf("\n"); ++cnt;
    }
    else
    {
        for(int i = 0; i < 8; ++i)
        {
            if(isValid(row, i))
            {
                board[row][i] = 1;
                queen_rec(row+1);
                board[row][i] = 0;
            }
        }
    }
}

void queen_nonrec()
{
//use stack to hold the previous valid position of each row
    int stack[8], row;
    for(row = 0; row < 8; ++row) stack[row] = -1;
    row = 0;
    //the row variable will decrease by 1 whenever backtracking happens. 
    //Finally, the row variable will become -1 and the loop should end there. 
    while(row >= 0)
    {
        if(row == 8)
        {
            for(int i = 0; i < 8; ++i)
            {
                for(int j = 0; j < 8; ++j)
                    if(board[i][j]) printf("%2d",1);
                    else printf("%2d", 0);
                printf("\n");
            }
            printf("\n"); ++cnt;
            --row;
        }
        else
        {
            //recover current position to be valid
            if(stack[row] >= 0 && stack[row] < 8) board[row][stack[row]] = 0;
            //try to find the next position available
            ++stack[row];
            while(stack[row] < 8 && !isValid(row, stack[row])) ++stack[row];
            //when come to the end of current row, set the position index of current row back to -1 and go to the upper row
            if(stack[row] >= 8) {stack[row] = -1; --row;}
            else
            {
                //found a valid position, go to the next row
                board[row][stack[row]] = 1; ++row;
            }
        }
    }
}

转载于:https://www.cnblogs.com/segeon/archive/2012/10/06/2712777.html