Leetcode N-Queens系列

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#51 N-Queens

The n-queens puzzle is the problem of placing n queens on ann×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.."]]

经典的回溯问题，为了减少对每个方格是否可放置皇后的判断，提前用一个数组记录下来。

代码中 isvalid 不同下标范围具有不同的含义：

0 ~ n- 1 index = col 代表第 col 列的状态（共n列）

n ~ 3n-2 index = n + row + col 代表（row，col）所在的正斜线的状态（共2n -1个正斜线）

3n -1 ~ 5n -3 index = 3n - 2 + n -row + col 代表（row, col）所在反斜线的状态（共2n -1个反斜线）

class Solution {public:    vector< vector<string> > solveNQueens(int n) {        vector< vector<string> > res;        if (n == 0) return res;        int row = 0;        vector<int> isvalid(n + 2 * (2 * n - 1), 1);        vector<int> col(n, -1);        while(row >= 0){            if(col[row] != -1) {                isvalid[col[row]] = isvalid[n + row + col[row]] = isvalid[3 * n - 2 + n  - row + col[row]] = 1;            }            col[row] ++;            while(col[row] < n){                if(isvalid[col[row]] && isvalid[n + row + col[row]] && isvalid[3 * n - 2 + n - row + col[row]] ) break;                col[row] ++;            }            if(col[row] == n)  row--; //回溯            else if(row == n -1){                vector<string> one;                for(int i = 0; i < n; i ++){                    string s = "";                    s.insert(0, n - 1, '.');                    s.insert(col[i], 1, 'Q');                    one.push_back(s);                }                res.push_back(one);            }            else{                isvalid[col[row]] = isvalid[n + row + col[row]] = isvalid[3 * n - 2 + n - row + col[row]] = 0;                col[++row] = -1;           }        }        return res;    }};

#52 N-Queens II

Follow up for N-Queens problem.

Now, instead outputting board configurations, return the total number of distinct solutions.

由上面代码稍加修改即可：

class Solution {public:    int totalNQueens(int n) {        if (n == 0) return 0;        int res = 0, row = 0;        vector<int> isvalid(n + 2 * (2 * n - 1), 1);        vector<int> col(n, -1);        while(row >= 0){            if(col[row] != -1) {                isvalid[col[row]] = isvalid[n + row + col[row]] = isvalid[3 * n - 2 + n  - row + col[row]] = 1;            }            col[row] ++;            while(col[row] < n){                if(isvalid[col[row]] && isvalid[n + row + col[row]] && isvalid[3 * n - 2 + n - row + col[row]] ) break;                col[row] ++;            }            if(col[row] == n)  row--; //回溯            else if(row == n -1) res ++;            else{                isvalid[col[row]] = isvalid[n + row + col[row]] = isvalid[3 * n - 2 + n - row + col[row]] = 0;                col[++row] = -1;           }        }        return res;    }};

可以写成递归形式如下：

class Solution {public:    void dfs(int &res, vector<int> &isvalid, int row, int n) {        if (row == n){            ++res; return;        }        for (int col = 0; col < n; col++) {            if (isvalid[col] && isvalid[n + row + col] && isvalid[3 * n - 2 + n - row + col]){                isvalid[col] = isvalid[n + row + col] = isvalid[3 * n - 2 + n - row + col] = 0;                dfs(res, isvalid, row + 1, n);                isvalid[col] = isvalid[n + row + col] = isvalid[3 * n - 2 + n  - row + col] = 1;            }        }        return;    }    int totalNQueens(int n) {        if (n == 0) return 0;        int res = 0;        vector<int> isvalid(n + 2 * (2 * n - 1), 1);        dfs(res, isvalid, 0, n);        return res;    }};

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