leetcode51. N-Queens 【更新 加上leetcode52 N-Queens II

题目要求

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.."]
]

这里的王后相当于国际围棋的王后，该王后一共有三种移动方式：水平、垂直，对角线。问要如何将n个王后布局在n*n的棋盘中保证她们不会互相伤害捏？

思路一：利用Set和Map

利用动态编程的思想。只要当前列，以及左对角线和右对角线都没有被占用的话，就可以在当前位置上暂时填上一个值。并将当前的临时结果作为下一轮的输入进行下一轮的预判。
在这里我们用Map存储每一行填入Q的列数，用Set存储占据的对角线。

    public List> solveNQueens(int n) {
        List> result = new ArrayList>();
        if(n==0){
            return result;
        }
        //记录左右对角线情况
        Set leftCordinal = new HashSet(), rightCordinal = new HashSet();
        Map current = new HashMap();
        solveNQueens(result, current, leftCordinal, rightCordinal, n);
        return result;
    }
    
    public void solveNQueens(List> result, Map current, Set leftCordinal, Set rightCordinal, int n){
        if(current.size() == n){
            List currentResult = new ArrayList();
            StringBuilder s = new StringBuilder();
            for(int i = 0 ; i

在这里其实可以用boolean[]代替Set，用简单值类型的数组代替复杂的数据结构往往可以提高性能。

    public List> solveNQueens2(int n) {
        List> result = new ArrayList>();
        if(n==0){
            return result;
        }
        //记录左右对角线情况
        boolean[] leftCordinal = new boolean[n*2-1], rightCordinal = new boolean[n*2-1];
        Map current = new HashMap();
        solveNQueens2(result, current, leftCordinal, rightCordinal, n);
        return result;
    }
    
    public void solveNQueens2(List> result, Map current, boolean[] leftCordinal, boolean[] rightCordinal, int n){
        if(current.size() == n){
            List currentResult = new ArrayList();
            StringBuilder s = new StringBuilder();
            for(int i = 0 ; i

将Map类型也转化一下？

我们发现，每一次得到一个结果时，都需要将相应的值转化为String然后添加的数组再添加到结果集中。但是其实这些String 的值都是重复利用的。毕竟Q可能位于0~n-1的任何位置上，也就是说有n个固定的互异字符串，但是一个结果上一定会包含所有这些字符串。这里我们进行了进一步的优化，将Map转化为boolean[]+int[]，其中bool数组用来记录该列是否被占领，int数组用来记录该行占领了那一列。

public class NQueens_51 {
    //将map数据结构int[]+boolean[]+lines[]+
    List> res;
    boolean[] col, lslash, rslash;
    int[] p;
    int n;
    String[] lines;
    private void dfs(int i) {
        if(i == n) {
            List board = new ArrayList();
            for(int j = 0; j < n; j++) {
                board.add(lines[p[j]]);
            }
            res.add(board);
            return;
        }
        
        for(int j = 0; j < n; j++) {
            if(!col[j] && !lslash[i+j] && !rslash[j-i+n-1]) {
                col[j] = true;
                lslash[i+j] = true;
                rslash[j-i+n-1] = true;
                p[i] = j;
                dfs(i+1);
                col[j] = false;
                lslash[i+j] = false;
                rslash[j-i+n-1] = false;
            }
        }
    }
    public List> solveNQueens3(int n) {
        this.n = n;
        res = new ArrayList>();
        col = new boolean[n];
        lslash = new boolean[2*n-1];
        rslash = new boolean[2*n-1];
        p = new int[n];
        char[] line = new char[n];
        lines = new String[n];
        for(int i = 0; i < n; i++) {
            line[i] = '.';
        }
        for(int i = 0; i < n; i++) {
            line[i] = 'Q';
            lines[i] = String.copyValueOf(line);
            line[i] = '.';
        }
        dfs(0);
        return res;
    }

}

N-Queens II 题目和实现

Follow up for N-Queens problem.

Now, instead outputting board configurations, return the total number of distinct solutions.

相比于I，这里不要去获得具体的答案，只需要返回解答的数量即可。其实比I要简单。思路还是相同的，利用迭代实现自顶向下的动态编程
代码如下：

    public int totalNQueens(int n) {
        boolean[] leftCordinal = new boolean[n*2-1], 
                  rightCordinal= new boolean[n*2-1], 
                  vertical = new boolean[n];
        int result = 0;
        return totalNQueens(0, n, result, vertical, leftCordinal, rightCordinal);
    }
    
    public int totalNQueens(int current, int n, int result, boolean[] vertical, boolean[] leftCordinal, boolean[] rightCordinal){
        if(current == n){
            return ++result;
        }
        
        for(int i = 0 ; i

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