Tuesday, November 18, 2014

[LeetCode] N-Queens I, II

N-Queens I

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.."]
]

N-Queens II

Follow up for N-Queens problem.
Now, instead outputting board configurations, return the total number of distinct solutions.

思路：
经典8皇后问题的推广版n皇后问题。两题其实是一回事，I的难度反而更大一些。因为能解I得到所有解，必然能知道一共几个解从而能解II。同样也是类似DFS的backtracking问题。难点在于如何判断当前某一位置是否可以放皇后，需要通过之前所有放置过的皇后位置来判断。对已经放置的任意皇后，需要判断当前位置是否在同一行、列、对角线上这三个条件。
1. 逐行放置皇后：排除在同一行的可能。
2. 记录之前所放皇后的列坐标：col[i]=j表示第i行的皇后在第j列。这样在放置第i+1行时，只要保证col[i+1] != col[k], k=0...i 即可。
3. 对角线判断：对于任意(i1, col[i1])和(i2, col[i2])，只有当abs(i1-i2) = abs(col[i1]-col[i2])时，两皇后才在同一对角线。

 N-Queens I

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class Solution {
public:
    vector<vector<string> > solveNQueens(int n) {
        vector<vector<string>> allSol;
        vector<string> sol;
        vector<int> col;
        solveNQ(n, 0, col, sol, allSol);
        return allSol;
    }
    
    void solveNQ(int n, int irow, vector<int> &col, vector<string> &sol, vector<vector<string>> &allSol) {
        if(irow==n) {
            allSol.push_back(sol);
            return;
        }
        
        for(int icol=0; icol<n; icol++) {
            if(validPos(col, irow, icol)) {
                string s(n,'.');
                s[icol] = 'Q';
                sol.push_back(s);
                col.push_back(icol);
                solveNQ(n, irow+1, col, sol, allSol);
                sol.pop_back();
                col.pop_back();
            }
        }
    }
    
    bool validPos(vector<int> &col, int irow, int icol) {
        if(irow<col.size()) return false;
        for(int i=0; i<col.size(); i++) {
            if(icol==col[i] || abs(irow-i)==abs(icol-col[i]))
                return false;
        }
        return true;
    }
};


N-Queens II


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class Solution {
public:
    int totalNQueens(int n) {
        vector<int> col;
        int totSol = 0;
        solveNQ(n, 0, col, totSol);
        return totSol;
    }
    
    void solveNQ(int n, int irow, vector<int> &col, int &totSol) {
        if(irow==n) {
            totSol++;
            return;
        }
        
        for(int icol=0; icol<n; icol++) {
            if(validPos(col, irow, icol)) {
                col.push_back(icol);
                solveNQ(n, irow+1, col, totSol);
                col.pop_back();
            }
        }
    }
    
    bool validPos(vector<int> &col, int irow, int icol) {
        if(irow<col.size()) return false;
        for(int i=0; i<col.size(); i++) {
            if(icol==col[i] || abs(irow-i)==abs(icol-col[i]))
                return false;
        }
        return true;
    }    
};
at

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