leetcode 51. N-Queens

最新推荐文章于 2025-05-30 23:37:00 发布

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最新推荐文章于 2025-05-30 23:37:00 发布

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本文介绍了一种解决N皇后问题的有效算法。通过递归回溯的方式，在n×n的棋盘上放置n个皇后，使得任意两个皇后不会互相攻击。文章详细展示了如何检查每一步放置的可行性，并列举了所有可行的解决方案。

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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..”]
]

List<List<Integer>> results = new ArrayList<List<Integer>>();
    List<Integer> result = new ArrayList<Integer>();

    public List<List<String>> solveNQueens(int n) {
        List<List<String>> result = new ArrayList<List<String>>();
        solveNQueens(n, 0);
        char[] chars = new char[n]; 
        //for(char mychar:chars) mychar = '.';//**********!!!!!!!!!!
        for(int i = 0;i < n;i++) chars[i] = '.';
        for(List<Integer> integers:results){
            List<String> strings = new ArrayList<String>();
            for(Integer integer:integers){//1 3 0 2
                chars[integer] = 'Q';
                strings.add(new String(chars));
                chars[integer] = '.';
            }
            result.add(strings);
        }
        return result;
    }

    public void solveNQueens(int n,int currentLine){
        if(currentLine == n) {
            List<Integer> integers = new ArrayList<Integer>();
            integers.addAll(result);
            results.add(integers);
        }
        else{
            for(int i = 0;i < n;i++){
                boolean placeAble = true;
                if(currentLine != 0){
                    if(result.contains(i))
                        placeAble = false;
                    else
                        for(int j = 0;j < currentLine;j++){
                            if((Math.abs(result.get(j)-i)) == currentLine - j){
                                placeAble= false;
                                break;
                            }
                        }
                }
                if(placeAble == true)
                {   result.add(i);
                    solveNQueens(n,currentLine+1);
                    result.remove(currentLine);
                }
            }
        }
    }