[leetcode-51]N-Queens(java)

问题描述：
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..”]
]

分析：这道题是典型的回溯法（递归+剪枝），从第一行开始向下，一直到最后一行，如果都找到了，那么将其进行处理之后添加到res中。这里，我的tmpList存的是col值，为的是方便进行比对（同一行，同一列或同一斜行）。

代码如下：288ms

public class Solution {
     public List> solveNQueens(int n) {
        List> res = new LinkedList<>();
        List tmpList = new LinkedList<>();//store col

        solve(n,0,res,tmpList);
        return res;
    }
    private void solve(int n,int index,List> res,List tmpList){
        if(n==index){
            List tmpStrList = new LinkedList<>();
            for(int i = 0;iint col = tmpList.get(i);
                StringBuffer sb = new StringBuffer();
                for(int j = 0;jif(j==col)
                        sb.append('Q');
                    else
                        sb.append('.');
                }
                tmpStrList.add(sb.toString());
            }
            res.add(tmpStrList);
            return;
        }
        //这个地方是O(N2)的复杂度，居然没有超时，，
        for(int col = 0;colint row = index;
            int rowList;
            for(rowList = 0;rowListint rowrow = rowList;
                int colcol = tmpList.get(rowrow);
                //同一列
                if(col == colcol)
                    break;
                //同一斜线
                if(Math.abs(rowrow-row)==Math.abs(colcol-col))
                    break;
            }
           if(rowList==tmpList.size()) {
               tmpList.add(col);
               solve(n, index + 1, res, tmpList);
               tmpList.remove(tmpList.size()-1);
           }
        }
    }
}