N-Queens

        The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.

        Given an integer n, return all distinct solutions to the n-queens puzzle.

        Each solution contains a distinct board configuration of the n-queens'placement, where 'Q' and '.' both indicate a queen and an empty space respectively.

        For example,
There exist two distinct solutions to the 4-queens puzzle:

[

 [".Q..",  // Solution 1

 "...Q",

 "Q...",

 "..Q."],

 

 ["..Q.",  // Solution 2

 "Q...",

 "...Q",

 ".Q.."]

]

 

        思路：本题又是一个典型的回溯法题型，采用递归方法解决。在第i行找到一个有效的位置后，接着递归深入到第i+1行寻找有效的位置，如果能找到，则继续深入；如果找不到，则回退到第i行，继续寻找第i行另一个有效位置。只要在最后一行找到一个有效位置，则表明找到了一个合格的答案。然后接着寻找最后一行下一个有效位置即可。

        因为是以行数从上到下进行深入的，所以在判断有效位置的时候，只需要判断到当前行即可。

        返回结果需要放到一个数组中，每个数组元素就是一个“棋盘”。因事先无法预料有多少个解法，需要边寻找边计数，所以本算法首先将所有结果放到一个链表中，然后在将链表复制到数组中。代码如下：

 

typedef struct queennode{    char **board;    struct queennode *next;}QueenNode; int isvalidqueen(char**board, int line, int col, int n){    int i, j;     for(i = 0; i <line; i++)    {        if(board[i][col]== 'Q')    return 0;    }     for(i = line-1,j = col-1; i>=0 && j>=0;i--,j--)    {        if(board[i][j]== 'Q')  return 0;    }     for(i = line-1,j = col+1; i>=0 && j<n;i--,j++)    {        if(board[i][j]== 'Q')  return 0;    }     return 1;} QueenNode *copyboard(char**board, int n){    int i;    QueenNode *res = calloc(1,sizeof(QueenNode));    res->board = calloc(n,sizeof(char *));     for(i = 0; i <n; i++)    {        res->board[i] =calloc(n, sizeof(char));        memcpy(res->board[i],board[i], n*sizeof(char));    }    return res;} void nqueencore(char**board, int line, int n, QueenNode **head,int *listlen){    int i;    QueenNode *res;    for(i = 0; i <n; i++)    {        if(isvalidqueen(board,line, i, n))        {            board[line][i]= 'Q';            if(line == n-1)            {                res = copyboard(board,n);                res->next = *head;                *head = res;                *listlen += 1;            }            else            {                nqueencore(board,line+1, n, head, listlen);            }              board[line][i]= '.';        }    }}  char*** solveNQueens(intn, int* returnSize){    char **board = calloc(n,sizeof(char *));    int i, j;    int reslen = 0;    char ***res;    for(i = 0; i <n; i++)    {        board[i] = calloc(n,sizeof(char));        memset(board[i], '.',n);    }     QueenNode *head = NULL;    QueenNode *p = NULL;    QueenNode *q = NULL;     nqueencore(board, 0, n,&head, &reslen);    res = calloc(reslen, sizeof(char**));     for(i = 0, p = head;i < reslen && p != NULL; i++)    {        res[i] = p->board;        q = p;        p = p->next;        free(q);    }    *returnSize = reslen;     return res;}