Leetcode N-Queens II

来源：互联网发布：英格拉姆体测数据编辑：程序博客网时间：2024/05/17 17:44

题意：要求输出N皇后问题的解的个数。

思路：返回解的个数。

class Solution {public:    vector<vector<string> > re;    int l;    int totalNQueens(int n) {        l = n;        string ts;        vector<string> tes;        for(int i = 0; i < n; i ++) ts += '.';        for(int i = 0; i < n; i ++) tes.push_back(ts);                dfs(tes, 0);                return re.size();    }    void dfs(vector<string> grid, int s) {        if(s == l) {            for(int i = 0; i < l; i ++) {                for(int j = 0; j < l; j ++) {                    if(grid[i][j] == 'x') grid[i][j] = '.';                }            }            re.push_back(grid);            return;        }                vector<string> mygrid;        for(int i = 0; i < l; ++ i) {            mygrid = grid;            if(mygrid[s][i] == 'x') continue;            if(mygrid[s][i] == '.') {                mygrid[s][i] = 'Q';                for(int j = s + 1; j < l; j ++) mygrid[j][i] = 'x';                for(int j = i - 1, k = s + 1; j >=0 && k < l; j --, k ++) mygrid[k][j] = 'x';                for(int j = i + 1, k = s + 1; j < l && k < l; j ++, k ++) mygrid[k][j] = 'x';                                                //for(int it = 0; it < l; it ++) cout << mygrid[it] << endl;                //cout << endl;                dfs(mygrid, s + 1);            }        }                return;    }};

另一种思路是不生成可行解，直接判断是否可行。判断是否可行有O(1)的方法。对于(x, y) 上的某个皇后，满足xi + yi = x + y和 n - xi + yi = x + y的位置都不能放置。

class Solution {public:    int count;    int l;    int totalNQueens(int n) {        vector<bool> c1(n, false);        vector<bool> c2(n * 2, false);        vector<bool> c3(n * 2, false);        count = 0;        l = n;                dfs(n, c1, c2, c3);                return count;    }        void dfs(int row, vector<bool> c1, vector<bool> c2, vector<bool> c3) {        if(row == 0) count ++;                for(int i = 0; i < l; ++ i) {            if(c1[i] || c2[row + i] || c3[l - row + i]) continue;            else {                c1[i] = true;                c2[row + i] = true;                c3[l - row + i] = true;                dfs(row - 1, c1, c2, c3);                c1[i] = false;                c2[row + i] = false;                c3[l - row + i] = false;            }        }    }};

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