• LeetCode 50 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:回溯法。使用ArrayList<Integer>  al来存储某个皇后的列,而它在ArrayList的索引就是其行数。
    public class Solution {
    	public List<String[]> solveNQueens(int n) {
    		List<String[]> result = new LinkedList<String[]>();
    		search(result, n, new ArrayList<Integer>(), 1);
    		return result;
    	}
    
    	private boolean isValid(List<Integer> list) {
    		int column = list.get(list.size() - 1);
    		for (int i = 0; i < list.size() - 1; i++) {
    			if (column == list.get(i)
    					|| Math.abs(column - list.get(i)) == Math.abs(list.size() - 1
    							- i)) {
    				return false;
    			}
    		}
    		return true;
    	}
    
    	private void search(List<String[]> result, int n, List<Integer> list,
    			int col) {
    		if (col > n) {
    			char[] temp = new char[n];
    			Arrays.fill(temp, '.');
    			String[] str = new String[n];
    			for (int i = 0; i < list.size(); i++) {
    				StringBuilder sb = new StringBuilder(String.valueOf(temp));
    				sb.setCharAt(list.get(i), 'Q');
    				str[i] = sb.toString();
    			}
    			result.add(str);
    		} else {
    			for (int j = 0; j < n; j++) {
    				list.add(j);
    				if (isValid(list)) {
    					search(result, n, list, col + 1);
    				}
    				list.remove(list.size() - 1);
    			}
    		}
    	}
    }
    思路2:回溯法,仅仅只是使用了位移操作,更快,剪枝剪得更厉害,具体分析见http://blog.csdn.net/mlweixiao/article/details/40984589
    public class Solution {
    	static int upperlim = 1;
    
    	private void search(int l, int ld, int rd, List<Integer> list,
    			List<String[]> result) {
    		if (l != upperlim) {
    			int pos = upperlim & (~(l | ld | rd));
    			while (pos != 0) {
    				int p = pos & -pos;
    				int temp = p;
    				int index = 0;
    				while (temp != 1) {
    					temp = temp >> 1;
    					index++;
    				}
    				list.add(index);// 从0開始的
    				pos ^= p;
    				search(l | p, (ld | p) << 1, (rd | p) >> 1, list, result);
    				list.remove(list.size() - 1);
    			}
    		} else {
    			char[] temp = new char[list.size()];
    			Arrays.fill(temp, '.');
    			String[] str = new String[list.size()];
    			for (int i = 0; i < list.size(); i++) {
    				StringBuilder sb = new StringBuilder(String.valueOf(temp));
    				sb.setCharAt(list.get(i), 'Q');
    				str[i] = sb.toString();
    			}
    			result.add(str);
    		}
    	}
    
    	public List<String[]> solveNQueens(int n) {
    		upperlim = (1 << n) - 1;
    		List<String[]> result = new LinkedList<String[]>();
    		search(0, 0, 0, new LinkedList<Integer>(), result);
    		return result;
    	}
    }



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  • 原文地址:https://www.cnblogs.com/yjbjingcha/p/7138010.html
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